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Precalculus : Exponential and Logarithmic Functions

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Example Questions

Example Question #13 : Rational Exponents

Evaluate the following expression and solve for .

$343=7^{15t-27}$

Possible Answers:

$t=\sqrt{27}$

t=-2

t=7

t=2

t=3

Correct answer:

t=2

Explanation:

To solve this problem, recall that you can set exponents equal to eachother if they have the same base.

See below:

343=7^3

So, we have

$7^3=7^{15t-27}$

Because both sides of this equation have a base of seven, we can set the exponents equal to eachother and solve for t.

3=15t-27

30=15t

t=2

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Example Question #14 : Rational Exponents

Solve for .

$2^{x/3}\cdot 3^{x/4}\cdot e^{-x}=3$

Possible Answers:

$x=\frac{3}{\ln432e^{-12}}\approx-0.506$

$x=\frac{36}{\ln432e^{-12}}\approx-6.069$

$x=\frac{36}{\ln432}-1\approx4.932$

$x=\frac{\ln3}{\ln(2^{1/3}\cdot3^{1/4}\cdot e^{-1})}\approx-2.223$

Correct answer:

$x=\frac{\ln3}{\ln(2^{1/3}\cdot3^{1/4}\cdot e^{-1})}\approx-2.223$

Explanation:

We begin by taking the natural log of the equation:

$\ln(2^{x/3}\cdot 3^{x/4}\cdot e^{-x})=\ln3$

Simplifying the left side of the equation using the rules of logarithms gives:

$\frac{x}{3}\ln2+\frac{x}{4}\ln3-x=\ln3$

We group the x terms to get:

$x\left(\frac{1}{3}\ln2+\frac{1}{4}\ln3-1\right)=\ln3$

We reincorporate the exponents into the logarithms and use the identity property of the natural log to obtain:

$x(\ln2^{1/3}+\ln3^{1/4}+\ln e^{-1})=\ln3$

We combine the logarithms using the multiplication/sum rule to get:

$x\ln(2^{1/3} 3^{1/4}e^{-1})=\ln3$

We then solve for x:

$x=\frac{\ln3}{\ln(2^{1/3}3^{1/4}e^{-1})}\approx-2.223$

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Example Question #1156 : Pre Calculus

Solve for .

$x^{5/2}-12x^{3/2}+20x^{1/2}=0$

Possible Answers:

x=2,10

x=0

x=0,4,5

x=0,2,10

Correct answer:

x=0,2,10

Explanation:

We begin by factoring out the term $x^{1/2}$ to get:

$x^{1/2}(x^2-12x+20)=0$

This equation gives our first solution:

$x^{1/2}=0\Rightarrow x=0$

Then we check for more solutions:

$\\x^2-12x+20=0 \\ \Rightarrow (x-10)(x-2)=0\\ \Rightarrow x=2,10$

Therefore our solution is

x=0,2,10

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Example Question #13 : Simplify Expressions With Rational Exponents

Evaluate $\frac{3xz^{\frac{1}{3}}}{(6x)^{\frac{1}{2}}y^{2}z^{3}}$ when $x=2,y=3^{\frac{1}{4}}, z=8$

Possible Answers:

$\frac{1}{256}$

$\frac{\sqrt{3}}{8^{8}}$

$\frac{3\sqrt{8}}{2^{8}\sqrt{6}}$

$\frac{\sqrt{3}}{ 2^{8}\sqrt{2}}$

Correct answer:

$\frac{1}{256}$

Explanation:

Remember the denominator of a rational exponent is equivalent to the index of a root.

This should simplify quite nicely.

$\frac{3xz^{\frac{1}{3}}}{(6x)^{\frac{1}{2}}y^{2}z^{3}}=\frac{3x^{1/2}}{\sqrt{2\cdot3}y^2z^{8/3}}=\frac{\sqrt{3x}}{\sqrt{2}y^2\sqrt[3]{z^{8}}}$

When $x=2,y=3^{\frac{1}{4}}, z=8$ it gives us,

$\frac{\sqrt{3\cdot2}}{\sqrt{2}(3^{\frac{1}{4}})^2\sqrt[3]{8^{8}}}=\frac{\sqrt{3}}{(3^{\frac{2}{4}})2^{8}}=\frac{1}{2^8}=\frac{1}{256}$

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Example Question #1158 : Pre Calculus

What is the value of $125^\frac{1}{3}$ ?

Possible Answers:

41.7

122

Correct answer:

Explanation:

What does an exponent of one-third mean? Consider our expression and raise it to the third power.

$(125^\frac{1}{3})^3 = x^3$

Simplifying, we get:

125 = x^3

Thus, we are looking for a number that when cubed, we get 125 . Thus, we are discussing the cube root of 125 , or .

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Example Question #1 : Solve Logarithmic Equations

Evaluate a logarithm.

What is $log_{10}1000$ ?

Possible Answers:

100

$\frac{1}{3}$

Correct answer:

Explanation:

The derifintion of logarithm is:

$log_{a}y=x \Leftrightarrow a^{x}=y.$

In this problem,

a= 10, y=1000

$10^{3}=1000$

Therefore, x=3

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Example Question #1 : Solve Logarithmic Equations

Solve for in the following logarithmic equation:

$log_{10} x^2 - log_{10} 2x = 2$

Possible Answers:

x = 10

x = 1000

None of the other choices

x=100

x = 0

Correct answer:

None of the other choices

Explanation:

Using the rules of logarithms,

$\log a - \log b = \log \bigg(\frac{a}{b}\bigg)$

Hence,

$\log x^2 - \log 2x = log \bigg(\frac{x^2}{2x}\bigg) = log \bigg(\frac{x}{2}\bigg) = 2$

So exponentiate both sides with a base 10:

$10^{log_{10} \frac{x}{2}} = 10^{2}$

The exponent and the logarithm cancel out, leaving:

$\frac{x}{2}=10^2$

$x=100\cdot 2=200$

This answer does not match any of the answer choices, therefore the answer is 'None of the other choices'.

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Example Question #1 : Solve Logarithmic Equations

Solve the following logarithmic equation:

$2\log _3(x)=\log _3(4)+\log _3(x-1)$

Possible Answers:

x=4

x=1

x=-4

x=2

x=-1

Correct answer:

x=2

Explanation:

In order to solve this equation, we must apply several properties of logarithms. First we notice the term on the left side of the equation, which we can rewrite using the following property:

$a\log_b (x)=\log_b (x^a)$

Where a is the coefficient of the logarithm and b is some arbitrary base. Next we look at the right side of the equation, which we can rewrite using the following property for the addition of logarithms:

$\log _ba+\log _bc=\log_b (a\cdot c)$

Using both of these properties, we can rewrite the logarithmic equation as follows:

$2\log _3(x)=\log _3(4)+\log _3(x-1)$

$\log _3(x^2)=\log _3(4(x-1))$

We have the same value for the base of the logarithm on each side, so the equation then simplifies to the following:

$x^2=4x-4\rightarrow x^2-4x+4=0$

Which we can then factor to solve for :

$(x-2)(x-2)=0\rightarrow x=2$

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Example Question #2 : Solve Logarithmic Equations

Solve the equation for .

$\ln\frac{cx}{ab+c} = b+c$

Possible Answers:

$x = \frac{e^b}{c(ab+c)e^c}$

$x = \frac{e^be^c(ab+c)}{c}}$

None of the other answers.

$x = \frac{e^a}{e^{bc}}}(ab+c)$

ab-e^c

Correct answer:

$x = \frac{e^be^c(ab+c)}{c}}$

Explanation:

We solve the equation as follows:

$\ln \frac{cx}{ab+c} = b+c$

Exponentiate both sides.

$\frac{cx}{ab+c}= e^{b+c}$

Apply the power rule on the right hand side.

$\frac{cx}{ab+c} = e^be^c$

Multiply by ab+c .

cx = e^be^c(ab+c)

Divide by .

$x =\frac{e^be^c(ab+c)}{c}$

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Example Question #4 : Solve Logarithmic Equations

Solve for :

log(27)- 2log(x)=log(x)

Possible Answers:

x=54

x=3

x=-3

$x=-\frac{1}{27}$

x=3.67

Correct answer:

x=3

Explanation:

First, simplify the logarithmic expressions on the left side of the equation:

2log(x) can be re-written as log(x^2) .

Now we have:

log(27)-log(x^2)=log(x) .

The left can be consolidated into one log expression using the subtraction rule:

$log\left(\frac{27}{x^2}\right)=log(x)$ .

We now have log on both sides, so we can be confident that whatever is inside these functions is equal:

$\frac{27}{x^2}=x$ to continue solving, multiply by x^2 on both sides:

27 = x^3 take the cube root:

3 = x

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Precalculus : Exponential and Logarithmic Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #13 : Rational Exponents

Example Question #14 : Rational Exponents

Example Question #1156 : Pre Calculus

Example Question #13 : Simplify Expressions With Rational Exponents

Example Question #1158 : Pre Calculus

Example Question #1 : Solve Logarithmic Equations

Example Question #1 : Solve Logarithmic Equations

Example Question #1 : Solve Logarithmic Equations

Example Question #2 : Solve Logarithmic Equations

Example Question #4 : Solve Logarithmic Equations

All Precalculus Resources

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