Int-Alg Chapter 7 Exponential Functions

Section A.7 Chapter 7 Exponential Functions

Subsection A.7.1 Exponential Growth and Decay

Subsubsection A.7.1.1 Compute percent increase and decrease

To calculate an increase of

r

%, we write the percent as a decimal and multiply the old amount by

1 + r .

To calculate a decrease we multiply the old amount by

1 - r .

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Example A.7.1.

A loaf of bread cost $3.00 last month, but this year the price rose by 6%. What should you multiply by to find the new price? What is the new price?

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Solution.

To get the new price, we multiply by 1.06 to get

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1.06 (3.00) = 3.18

The new price is $3.18.

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Example A.7.2.

Priceco is offering a 15% discount off the regular price of $180 for a ceiling fan. What should you multiply by to find the new price? What is the new price.

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Solution.

To get the new price, we multiply by

1 - 0.15,

or 0.85, to get

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0.85 (180) = 153

so the new price is $153.

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Checkpoint A.7.3.

Muriel’s rent was increased by 8% from $650 per month. What should you multiply by to find her new rent? What is her new rent?

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Answer.

1.08,

$702

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Checkpoint A.7.4.

A brand new SUV loses 18% of its value as soon as you drive it off the lot. If your SUV cost $35,000, what should you multiply to find its new value? What is its new value?

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Answer.

0.82,

$28,700

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Subsubsection A.7.1.2 Use the order of operations

Recall that evaluating powers comes before multiplication in the order of operations.

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Example A.7.5.

Simplify.

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$- 4 - 2^{3}$
$- 4 (- 2)^{3}$
$(- 4 - 2)^{3}$

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Solution.

Compute $2^{3}$ first, then subtract the result from $- 4 :$
$- 4 - 2^{3} = - 4 - 8 = - 12$
Compute $(- 2)^{3}$ first, then multiply the result by $- 4 :$
$- 4 (- 2)^{3} = - 4 (- 8) = 32$
Compute $(- 4 - 2)$ first, then cube the result:
$(- 4 - 2)^{3} = (- 6)^{3} = - 216$

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Example A.7.6.

Evaluate for

x = 6 .

Round your answers to hundredths.

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$12 (1.05)^{x}$
$12 (1 + x / 100)^{5}$

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Solution.

Follow the order of operations. Compute the power first:

$12 (1.05)^{6} = 12 (1.3400956 . . .) = 16.08$
Follow the order of operations. Compute the power first:

$12 (1 + 6 / 100)^{5} = 12 (1.06)^{5} = 12 (1.3382255 . . .) = 16.06$

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Checkpoint A.7.7.

Simplify. Round your answers to the nearest whole number.

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$450 (1 - 0.12)^{4}$
$180 - 80 (1 + 0.25)^{3}$

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Answer.

$270$
$24$

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Checkpoint A.7.8.

Evaluate for

x = - 3, y = - 2 .

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$- 2 x^{2} + y^{3}$
$4 (x - y) (x + 2 y)$

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Answer.

$- 26$
$28$

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Subsubsection A.7.1.3 Simplify expressions

Be careful to avoid tempting but false operations with exponents.

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Example A.7.9.

Which equation is a correct application of the laws of exponents?

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$20 (1 + r)^{4} = 20 + 20 r^{4}$ or $(a b^{t})^{3} = a^{3} b^{3 t}$
$2^{t / 5} = (2^{1 / 5})^{t}$ or $6.8 (10)^{t} = 68^{t}$

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Solution.

The first statement is not correct. There is no law that says $(a + b)^{n}$ is equivalent to $a^{n} + b^{n},$ so $(1 + r)^{4}$ is not equivalent to $1^{4} + r^{4}$ or $1 + r^{4} .$

However, it is true that $(a b)^{n} = a^{n} b^{n},$ so in particular the second statement is true:

$(a b^{t})^{3} = a^{3} (b^{t})^{3} = a^{3} b^{3 t}$
The first statement is correct. If we start with $(2^{1 / 5})^{t},$ we can apply the third law, $(a^{m})^{n} = a^{m n},$ to find

$(2^{1 / 5})^{t} = 2^{(1 / 5) t} = 2^{t / 5} .$

In the second statement, 6.8 is not raised to power $t,$ so we cannot multiply 6.8 times 10.

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Decide whether each simplification is a correct application of the laws of exponents. Write a correct statement if possible.

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Checkpoint A.7.10.

P (1 - r)^{6} \to P - P r^{6}

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Answer.

Not correct

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Checkpoint A.7.11.

25 (2^{t}) \cdot 4 (2^{t}) \to 100 \cdot 2^{t^{2}}

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Answer.

Not correct.

25 (2^{t}) \cdot 4 (2^{t}) = 100 (2^{2 t})

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Checkpoint A.7.12.

a {(b^{1 / 8})}^{2 t} \to a b^{t / 4}

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Answer.

Correct

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Checkpoint A.7.13.

N (0.94)^{1 / 8.3} \to \frac{N}{(0.94)^{8.3}}

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Answer.

Not correct, but

N (0.94)^{- 8.3} = \frac{N}{(0.94)^{8.3}}

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Subsection A.7.2 Exponential Functions

Subsubsection A.7.2.1 Evaluate exponential functions

Powers come before products in the order of operations, so to evaluate an exponential function

f (x) = a b^{x}

we evaluate

b^{x}

before multiplying by

a .

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Example A.7.14.

Evaluate

f (x) = 8 \cdot 4^{x} .

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$f (2)$
$f (- 2)$
$f (\frac{1}{2})$
$f (- \frac{1}{2})$

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Solution.

Follow the order of operations: compute powers before products.

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$f (2) = 8 \cdot 4^{2} = 8 \cdot 16 = 128$
$f (- 2) = 8 \cdot 4^{- 2} = 8 \cdot \frac{1}{16} = \frac{1}{2}$
$f (\frac{1}{2}) = 8 \cdot 4^{1 / 2} = 8 \cdot 2 = 16$
$f (- \frac{1}{2}) = 8 \cdot 4^{- 1 / 2} = 8 \cdot \frac{1}{2} = 4$

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Example A.7.15.

Evaluate

g (x) = 120 (0.65)^{x} .

Round your answers to thousandths.

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$g (2.3)$
$g (- 1.8)$
$g (0.4)$
$g (- 0.25)$

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Solution.

Follow the order of operations: use your calculator to compute powers before products. Do not round off at intermediate steps!

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$g (2.3) = 120 (0.65)^{2.3} = 120 (0.37127 . . .) = 44.554$
$g (- 1.8) = 120 (0.65)^{- 1.8} = 120 (2.17148 . . .) = 260.578$
$g (0.4) = 120 (0.65)^{0.4} = 120 (0.84171 . . .) = 101.006$
$g (- 0.25) = 120 (0.65)^{- 0.25} = 120 (1.11370 . . .) = 133.645$

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Checkpoint A.7.16.

Evaluate each function. Give your answers as common fractions.

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$G (t) = 15 (5)^{t} .$ Find $G (- 3)$
$H (n) = 4 {(\frac{1}{27})}^{n} .$ Find $H (\frac{2}{3})$
$F (x) = \frac{1}{2} \cdot 8^{x} .$ Find $F (- \frac{1}{3})$

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Answer.

$\frac{3}{25}$
$\frac{4}{9}$
$\frac{1}{4}$

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Checkpoint A.7.17.

Evaluate each function. Round your answers to hundredths.

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$G (t) = 15 (1.5)^{t} .$ Find $G (- 3)$
$h (z) = 1.8 (0.8)^{z} .$ Find $h (4)$
$F (w) = 2500 (1.03)^{w} .$ Find $F (25)$

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Answer.

$4.44$
$0.745$
$5234.44$

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Subsubsection A.7.2.2 Find an unknown exponent

If we can write both sides of an equation as powers with the same base, we can equate the exponents.

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Example A.7.18.

Find the value of the exponent.

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$3^{x} = 81$
$5^{x} = \frac{1}{125}$
${(\frac{3}{4})}^{x} = \frac{16}{9}$
$64^{x} = 16$

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Solution.

We can write both sides with base 3.

$81 = 3^{4}, so x = 4 .$
We can write both sides with base 5.

$125 = 5^{3}, so 5^{- 3} = \frac{1}{125}, and x = - 3 .$
${(\frac{3}{4})}^{2} = \frac{9}{16},$ so ${(\frac{3}{4})}^{- 2} = \frac{16}{9},$ and $x = - 2 .$
We can write both sides with base 4.

$64 = 4^{3} and 16 = 4^{2},$

so

$\begin{aligned} (4^{3})^{x} & = 4^{2} & Multiply exponents. \\ 4^{3 x} & = 4^{2} & Equate exponents. \\ 3 x & = 2 \\ x & = \frac{2}{3} \end{aligned}$

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Example A.7.19.

By using trial and error, estimate the value of the exponent to the nearest tenth.

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$2^{x} = 15$
$3^{x} = 65$
$10^{x} = 0.03$
${0.5}^{x} = 0.20$

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Solution.

$2^{4} = 16,$ so we try a slightly smaller exponent and find that $2^{3.9} = 14.9285,$ so $x \approx 3.9 .$
65 is between $3^{3} = 27$ and $3^{4} = 81,$ so $x$ must be between 3 and 4. By trying exponents 3.1, 3.2, 3.3, and so on, we find that $3^{3.8} = 65.022,$ so $x \approx 3.8 .$
$10^{- 1} = 0.1$ and $10^{- 2} = 0.01,$ so $- 2 < x < - 1 .$ By trying exponents between $- 2$ and $- 1,$ we find that $10^{- 1.5} = 0.0316,$ so $x \approx - 1.5 .$
${0.5}^{2} = 0.25,$ and as we increase the exponent on $0.5,$ the result will be smaller. By trial and error we find that ${0.5}^{2.3} = 0.2031,$ so $x \approx 2.3 .$

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Checkpoint A.7.20.

Find the value of the exponent.

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$2^{x} = \frac{1}{1024}$
$125^{x} = 25$

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Answer.

$10$
$\frac{2}{3}$

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Checkpoint A.7.21.

By using trial and error, estimate the value of the exponent to the nearest tenth.

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$10^{x} = 50$
${1.08}^{x} = 1.5$

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Answer.

$1.7$
$5.3$

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Subsubsection A.7.2.3 Solve equations graphically

We first solve equations graphically in Section 1.2, so you might want to review that subsection.

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Example A.7.22.

Here is a graph of

f (x) = x^{2} + 2 x - 16 .

Use the graph to solve the equation

x^{2} + 2 x - 16 = 8 .

Show your work on the graph.

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$parabola$

Solution.

To solve the equation, we want to find

x

-values that produce a function value of 8. The vertical coordinate of each point on the graph is given by the function value,

f (x) .

So we look for points on the graph with vertical coordinate

f (x) = 8 .

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$parabola with annotations$

There are two such points,

(- 6, 8)

and

(4, 8) .

Those points tell us that

f (- 6) = 8

and

f (4) = 8 .

Thus, the

x

-coordinates of the points, namely

- 6

and

4,

are the solutions. To check algebraically, we can verify that

f (- 6) = 8

and

f (4) = 8 :

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\begin{aligned} f (- 6) & = (- 6)^{2} + 2 (- 6) - 16 = 36 - 12 - 16 = 8 \\ f (4) & = 4^{2} + 2 (4) - 16 = 16 + 8 - 16 = 8 \end{aligned}

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Example A.7.23.

Here is a graph of

G (x) = \frac{20}{(x - 3)^{2}} .

Use the graph to solve the equation

\frac{20}{(x - 3)^{2}} = 5 .

Show your work on the graph.

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$transformed inverse-square$

Solution.

We find any points on the graph with vertical coordinate

G (x) = 5 .

There are two points,

(1, 5)

and

(5, 5 .

) The

x

-coordinates of those points, namely 1 and 5, are the solutions.

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$transformed inverse-square-annotated$

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Checkpoint A.7.24.

Here is a graph of

F (x) = \frac{1}{16} x^{3} + \frac{3}{8} x^{2} - \frac{3}{2} - 7

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$cubic$

Use the graph to solve the equation

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\frac{1}{16} x^{3} + \frac{3}{8} x^{2} - \frac{3}{2} - 7 = - 3

Show your work on the graph.

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Answer.

- 8,

- 2,

4

$cubic$

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Checkpoint A.7.25.

Here is a graph of

g (x) = 15 x^{2 / 3}

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$y=15 x to 2/3$

Use the graph to solve the equation

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15 x^{2 / 3} = 60

Show your work on the graph.

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Answer.

- 8, 8

$y=15 x to 2/3$

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Subsection A.7.3 Logarithms

Subsubsection A.7.3.1 Convert between radicals and powers

Because a logarithm is an exponent, it will be helpful to convert easily between radical notation and exponent notation.

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Example A.7.26.

Write each power as a radical.

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$x^{2 / 3}$
$t^{- 3 / 2}$
$w^{1.25}$
$z^{- 0.4}$

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Solution.

Recall that the numerator of the exponent is the power and the denominator is the root. A negative exponent indicates a reciprocal.

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$x^{2 / 3} = \sqrt[3]{x^{2}}$
$t^{- 3 / 2} = \frac{1}{t^{3 / 2}} = \frac{1}{\sqrt{t^{3}}}$
$w^{1.25} = w^{5 / 4} = \sqrt[4]{w^{5}}$
$z^{- 0.4} = \frac{1}{z^{4 / 10}} = \frac{1}{z^{2 / 5}} = \frac{1}{\sqrt[5]{z^{2}}}$

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Example A.7.27.

Write each radical expression in exponential form and simplify.

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$\sqrt[4]{b^{3}}$
$\frac{1}{\sqrt[6]{a^{3}}}$
$x^{2} \sqrt[4]{x}$
$\frac{\sqrt[3]{v}}{\sqrt{v}}$

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Solution.

$\sqrt[4]{b^{3}} = b^{3 / 4}$
$\frac{1}{\sqrt[6]{a^{3}}} = \frac{1}{a^{3 / 6}} = a^{- 3 / 6} = a^{- 1 / 2}$
$x^{2} \sqrt[4]{x} = x^{2} x^{1 / 4} = a^{2 + 1 / 4} = a^{9 / 4}$
$\frac{\sqrt[3]{v}}{\sqrt{v}} = \frac{v^{1 / 3}}{v^{1 / 2}} = v^{1 / 3 - 1 / 2} = v^{- 1 / 6}$

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Checkpoint A.7.28.

Write each power as a radical.

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$m^{- 3 / 5}$
$p^{2.75}$
$x^{0.18}$

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Answer.

$\frac{1}{\sqrt[5]{m^{3}}}$
$\sqrt[4]{p^{11}}$
$\sqrt[50]{x^{9}}$

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Checkpoint A.7.29.

Write each radical expression in exponential form and simplify.

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$\sqrt[10]{n^{9}}$
$\sqrt{h} \sqrt[4]{h}$
${(\sqrt[3]{t^{2}})}^{4}$

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Answer.

$n^{9 / 10}$
$h^{3 / 4}$
$t^{8 / 3}$

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Subsubsection A.7.3.2 Estimate logs

It is useful to be able to estimate mentally the value of a log.

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Example A.7.30.

Write each log equation in exponential form. Then use trial and error to estimate the log between two integers.

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$\log_{2} (6) = x$
$\log_{2} (24) = x$
$\log_{2} (100) = x$
$\log_{2} (0.3) = x$

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Solution.

$2^{x} = 6.$ Because $2^{2} = 4$ and $2^{3} = 8,$ $2 < x < 3 .$
$2^{x} = 24.$ Because $2^{4} = 16$ and $2^{5} = 32,$ $4 < x < 5 .$
$2^{x} = 100.$ Because $2^{6} = 64$ and $2^{7} = 128,$ $6 < x < 7 .$
$2^{x} = 0.3 .$ Because $2^{- 2} = \frac{1}{4} = 0.25$ and $2^{- 1} = \frac{1}{2} = 0.5,$ $- 2 < x < - 1 .$

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Example A.7.31.

Use computing technology to complete the table for $f (x) = 5^{x} .$ Round the function values to tenths.

$x$ $2.0$ $2.1$ $2.2$ $2.3$ $2.4$ $2.5$ $2.6$ $2.7$ $2.8$ $2.9$ $3.0$

$f (x)$

Use your table from part (a) to make a table of values for the function

g (x) = \log_{5} (x) .

$x$
$g (x)$

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Solution.

$x$ $2.0$ $2.1$ $2.2$ $2.3$ $2.4$ $2.5$ $2.6$ $2.7$ $2.8$ $2.9$ $3.0$

$f (x)$ $25$ $29.4$ $34.5$ $40.5$ $47.6$ $55.9$ $65.7$ $77.1$ $90.6$ $106.4$ $125$
$x$ $25$ $29.4$ $34.5$ $40.5$ $47.6$ $55.9$ $65.7$ $77.1$ $90.6$ $106.4$ $125$

$g (x)$ $2.0$ $2.1$ $2.2$ $2.3$ $2.4$ $2.5$ $2.6$ $2.7$ $2.8$ $2.9$ $3.0$

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Checkpoint A.7.32.

Write each log equation in exponential form. Then use trial and error to estimate the log, first between two integers, and then to the nearest tenth.

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$\log_{3} (10) = x$
$\log_{3} (20) = x$
$\log_{3} (150) = x$
$\log_{3} (0.5) = x$

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Answer.

$3^{x} = 10,$ between 2 and 3, $2.1$
$3^{x} = 20,$ between 2 and 3, $2.7$
$3^{x} = 150,$ between 4 and 5, $4.6$
$3^{x} = 0.5,$ between $- 1$ and 0, $- 0.6$

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Checkpoint A.7.33.

Use computing technology to complete the table for $f (x) = 4^{x} .$

$x$ $- 1$ $- 0.5$ $0$ $0.5$ $1$ $1.5$ $2$ $2.5$ $3$ $3.5$ $4.0$

$f (x)$

Use your table from part (a) to make a table of values for the function

g (x) = \log_{4} (x) .

$x$
$g (x)$

Use your table from part (b) to make a graph of $g (x) = \log_{4} (x) .$

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Answer.

$x$ $- 1$ $- 0.5$ $0$ $0.5$ $1$ $1.5$ $2$ $2.5$ $3$ $3.5$ $4.0$

$f (x)$ $0.25$ $0.5$ $1$ $2$ $4$ $8$ $16$ $32$ $64$ $128$ $256$
$x$ $0.25$ $0.5$ $1$ $2$ $4$ $8$ $16$ $32$ $64$ $128$ $256$

$g (x)$ $- 1$ $- 0.5$ $0$ $0.5$ $1$ $1.5$ $2$ $2.5$ $3$ $3.5$ $4.0$

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Subsubsection A.7.3.3 Using logs

Example A.7.34.

Solve each exponential equation by writing your answer as a logarithm, then evaluate if possible (without a calculator).

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$3 \cdot 2^{x} = 96$
$3 \cdot 2^{x} = 90$

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Solution.

First isolate the power to get $2^{x} = 32 .$ Then convert to logarithmic form as $x = \log_{2} (32) .$ Because $2^{5} = 32,$ we can simplify the answer to $x = 5 .$
First isolate the power to get $2^{x} = 30 .$ Then convert to logarithmic form as $x = \log_{2} (30) .$ This log is not easy to evaluate without a calculator.

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Checkpoint A.7.35.

Solve each exponential equation by writing your answer as a logarithm, then evaluate if possible (without a calculator).

🔗

$2 \cdot 5^{x} - 20 = 230$
$2 \cdot 5^{x} - 20 = 200$
$3^{x} = 81$
$3^{2 x - 5} = 81$
$10^{2 x} = 1000$
$10^{2 x} = 1200$
$5 + 4^{3 x} = 7$
$5 + 4^{3 x} = 8$

🔗

Answer.

$\log_{5} (125) = 3$
$\log_{5} (110)$
$\log_{3} (81) = 4$
$\frac{1}{2} [5 + \log_{3} (81)] = \frac{9}{2}$
$\frac{1}{3} \log_{4} (2) = \frac{1}{6}$
$\frac{1}{3} \log_{4} (3)$

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Subsection A.7.4 Properties of Logarithms

Let us review some of the algebraic properties we learned earllier as they apply to exponential functions.

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Subsubsection A.7.4.1 Apply the laws of exponents

The laws of exponents still apply to variable exponents. (If you would like to review the laws of exponents, they are listed in Section 3.2.)

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Example A.7.36.

Use the laws of exponents to simplify.

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${1.35}^{6} ({1.35}^{4})$
${0.64}^{5} ({0.64}^{n})$

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Solution.

When multiplying two powers with the same base, we add the exponents. Notice that the base does not change.

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${1.35}^{6} ({1.35}^{4}) = {1.35}^{6 + 4} = {1.35}^{10}$
${0.64}^{5} ({0.64}^{n}) = {0.64}^{5 + n}$

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Example A.7.37.

Use the laws of exponents to simplify.

🔗

$\frac{{0.32}^{8}}{{0.32}^{2}}$
$\frac{{0.32}^{t}}{{0.32}^{x}}$

🔗

Solution.

When dividing two powers with the same base, we subtract the exponents.

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$\frac{{0.32}^{8}}{{0.32}^{2}} = {0.32}^{8 - 2} = {0.32}^{6}$
$\frac{{0.32}^{t}}{{0.32}^{x}} = {0.32}^{t - x}$

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Example A.7.38.

Use the laws of exponents to simplify.

🔗

${({1.07}^{5})}^{3}$
${({1.07}^{4})}^{p}$

🔗

Solution.

When raising a power to a power, we multiply the exponents.

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${({1.07}^{5})}^{3} = {1.07}^{15}$
${({1.07}^{4})}^{p} = {1.07}^{4 p}$

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Checkpoint A.7.39.

Use the laws of exponents to simplify

{2.5}^{2 t} ({2.5}^{3}) .

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Answer.

{2.5}^{2 t + 3}

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Checkpoint A.7.40.

Use the laws of exponents to simplify

{({0.94}^{4})}^{m - 2} .

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Answer.

{0.94}^{4 m - 8}

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Checkpoint A.7.41.

Use the laws of exponents to simplify

\frac{{1.13}^{8 x}}{{1.13}^{5 x}} .

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Answer.

{1.13}^{3 x}

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Subsubsection A.7.4.2 Apply the distributive law

Example A.7.42.

Which equation is a correct application of the distributive law?

🔗

$2 (5 \cdot 3^{x}) = 10 \cdot 6^{x}$ or $2 (5 + 3^{x}) = 10 + 2 \cdot 3^{x}$
$\log (x + 10) = \log x + \log 10$ or $\frac{1}{x} (x + 10) = 1 + \frac{10}{x}$

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Solution.

The distributive law applies to multiplying a sum or difference, not a product. In the first equation, $5 \cdot 3^{x}$ is a product, so the distributive law does not apply. (We can, however, simplfy that expression with the associative law:

$2 (5 \cdot 3^{x}) = (2 \cdot 5) \cdot 3^{x} = 10 \cdot 3^{x}$

The second equation is a correct application of the distributive law. You can check that the first equation is false and the second equation is true by substituting $x = 1 .$
The distributive law applies only to multiplying a sum or product, not to other operations, such as taking logs. You can check that the first equation is false by substituting $x = 10 .$

The second equation is a correct application of the distributive law.

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Decide whether each equation is a correct application of the distributive law. Write a correct statement if possible.

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Checkpoint A.7.43.

\frac{x + 6}{3} \to \frac{x}{3} + \frac{6}{3}

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Answer.

Correct

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Checkpoint A.7.44.

\frac{6}{x + 3} \to \frac{6}{x} + \frac{6}{3}

🔗

Answer.

Not correct

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Checkpoint A.7.45.

2 (P_{0} a^{t}) \to 2 P_{0} + 2 a^{t}

🔗

Answer.

Not correct.

2 (P_{0} a^{t}) = 2 P_{0} a^{t}

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Checkpoint A.7.46.

25 (1 + r)^{8} \to (25 + 25 r)^{8}

🔗

Answer.

Not correct

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Subsubsection A.7.4.3 Solve power and exponential equations

Compare the procedures for solving power equations and exponential equations.

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Example A.7.47.

Solve

3 x^{1.05} = 18 .

Round your answer to hundredths.

🔗

Solution.

This is a power equation. We divide both sides by 3 to isolate the variable, then raise both sides to the reciprocal of the exponent.

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\begin{aligned} (x^{1.05})^{1 / 1.05} & = 6^{1 / 1.05} \\ x & = 5.51 \end{aligned}

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Example A.7.48.

Solve

3 (1.05)^{x} = 18 .

Round your answer to hundredths.

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Solution.

This is an exponential equation. We divide both sides by 3, then take logarithms.

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\begin{aligned} \log ({1.05}^{x}) & = \log 6 & Apply the third log property. \\ x \log 1.05 & = \log 6 \\ x & = \frac{\log 6}{\log 1.05} = 36.72 \end{aligned}

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Example A.7.49.

Solve

9 x^{3 / 5} = 36 .

Round your answer to hundredths.

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Solution.

This is a power equation. We divide both sides by 9 to isolate the variable, then raise both sides to the reciprocal of the exponent.

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\begin{aligned} (x^{3 / 5})^{5 / 3} & = 4^{5 / 3} \\ x & = 10.08 \end{aligned}

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Example A.7.50.

Solve

1.5 (3^{x / 5}) = 12 .

Round your answer to hundredths.

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Solution.

This is an exponential equation. We divide both sides by 1.5, then take logarithms.

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\begin{aligned} \log 3^{x / 5} & = \log 8 & Apply the third log property. \\ \frac{x}{5} \log 3 & = \log 8 \\ x & = \frac{5 \log 8}{\log 3} = 9.46 \end{aligned}

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Checkpoint A.7.51.

Solve

6 x^{3 / 4} - 8 = 76 .

Round your answer to hundredths.

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Answer.

33.74

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Checkpoint A.7.52.

Solve

6 {(\frac{3}{4})}^{x} - 8 = 76 .

Round your answer to hundredths.

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Answer.

- 9.17

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Checkpoint A.7.53.

Solve

13.2 (1.36)^{x} = 284.8 .

Round your answer to hundredths.

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Answer.

9.99

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Checkpoint A.7.54.

Solve

13.2 x^{1.26} = 284.8 .

Round your answer to hundredths.

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Answer.

11.45

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Subsection A.7.5 Exponential Models

Subsubsection A.7.5.1 Interpret function notation

The definitions of the variables help us interpret function notation.

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Example A.7.55.

The number of students at Salt Creek Elementary School is growing according to the formula

f (t) = 500 (1.08)^{t},

where

t

is the number of years since the school opened in 2005.

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What does the equation $f (6) = 500 (1.08)^{6}$ tell us about the school?
Use function notation to say that the student population was 583 in 2007.

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Solution.

In this equation, $t = 6$ and $f (6) = 793 .$ In 2011 (six years after the school opened), the student population was 793.
In 2007, $t = 2,$ so $f (2) = 500 (1.08)^{2} = 583 .$

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Example A.7.56.

The value of Digicorp stock has been falling according to the formula

V (w) = 48 (0.96)^{w},

where

w

is the number of weeks since its peak value of $48 per share.

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Use function notation to say that 8 weeks later the value of a share of Digicorp stock was $34.63.
What does the equation $V (12) = 48 (0.96)^{12} = 29.41$ tell us about the stock?

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Solution.

We evaluate the function at $w = 8$ to get $V (8) = 48 (0.96)^{8} = 34.63 .$
In this equation, $w = 12$ and $V (12) = 29.41,$ so 12 weeks after the peak value a share of Digicorp stock was worth $29.41.

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Checkpoint A.7.57.

The number of internet users in the United States is given by

I (t) = 95, 331, 000 (1.09)^{t},

where

t = 0

in 2000. Use function notation to say that the number of internet users in 2005 was 146,679,000.

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Answer.

I (t) = 146, 679, 000

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Checkpoint A.7.58.

The percent of U.S. households that maintain a landline telephone is decreasing according to the formula

L (t) = 95 (0.96)^{t},

where

t = 0

in 2004. What does the equation

L (t) = 95 (0.96)^{10} = 63

tell us about landlines?

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Answer.

In 2014, 63% of households maintained a landline.

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Subsubsection A.7.5.2 Doubling time and half-life

Doubling Time and Half-Life.

D

is the doubling time for an exponential function

P (t),

then

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P (t) = P_{0} 2^{t / D}

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H

is the half-life for an exponential function

Q (t),

then

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Q (t) = Q_{0} (0.5)^{t / H}

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Example A.7.59.

The half-life of a cold medication in the body is 6 hours. Find its decay rate.

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Solution.

The decay law for the medication is

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N = N_{0} (0.5)^{t / 8}

We can rewrite this expression as

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N = N_{0} ({0.5}^{1 / 8})^{t}

b = {0.5}^{1 / 8} = 0.9170,

and

r = 1 - b = 0.083 .

The decay rate is 8.3%.

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Example A.7.60.

The growth rate of a population of badgers is 3.8% per year. Find its doubling time.

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Solution.

The growth law for the population is

P = P_{0} (1.038)^{t} .

We set

P = 2 P_{0}

and solve for

t .

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\begin{aligned} 2 P_{0} & = P_{0} (1.038)^{t} & Divide both sides by P_{0} . \\ 2 & = (1.038)^{t} & Take the log of both sides. \\ \log 2 & = t \log 1.038 & Apply the third log property. \\ t & = \frac{\log 2}{\log 1.038} = 18.59 \end{aligned}

The doubling time is 18.59 years.

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Checkpoint A.7.61.

The doubling time for a population is 18 years. Find its annual growth rate.

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Answer.

3.9%

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Checkpoint A.7.62.

A radioactive isotope decays by 0.04% per second. What is its half-life?

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Answer.

4.81 hrs

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Subsubsection A.7.5.3 Analyze graphs of exponential functions

From a graph, we can read the initial value of an exponential function and then its doubling time or half-life. From there we can calculate the growth or decay law.

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Example A.7.63.

The graph shows the population,

P,

of a herd of llamas

t

years after 2000.

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How many llamas were there in 2000?
What is the doubling time for the population?
What is the annual growth rate for the population?

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$exponential growth$

Solution.

The initial value of the population is given by the $P$ -intercept of the graph, $(0, 15) .$ There were 15 llamas in 2000.
Look for the time when the initial llama population doubles. When $t = 4, P = 30,$ and when $t = 8, P = 60,$ so the llama population doubles every 4 years.
The growth factor for the population is $2^{1 / 4} = 1.189,$ so the annual growth rate is 18.9%.