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Precalculus : Exponential and Logarithmic Functions

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Example Questions

Example Question #31 : Exponential And Logarithmic Functions

Solve for x:

2^x=128

Possible Answers:

Correct answer:

Explanation:

2^x=128

2^x=2^7

$\log_22^x=\log_22^7$

x=7

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Example Question #32 : Exponential And Logarithmic Functions

Solve the equation:

$2^x4^2=8^{^{3+x}}$

Possible Answers:

$x=\frac{2}{5}$

$x=\frac{-5}{2}$

$x=\frac{-2}{5}$

$x=\frac{5}{2}$

x=-2

Correct answer:

$x=\frac{-5}{2}$

Explanation:

We notice that all of the bases are powers of 2 so we can re-write them as 2 raised to a certain power.

$2^x(2^2)^2=(2^3)^{3+x}$

Then we simplify both sides of the equation so we have a single base 2 raised to a power. When we have when we have exponentials with the same base multiplied we add the exponents, when we have them raised to an exponent we multiply the exponents.

$2^{x+4}=2^{9+3x}$

Now that we have the same base on both sides we set the exponents equal and solve for x.

x+4=9+3x

2x=-5

$x=\frac{-5}{2}$

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Example Question #32 : Exponential And Logarithmic Functions

The logarithms for this question are common logs (i.e. base 10).

Solve for in the following expression. Round to three decimal places.

$\log(x^2+10x+25) +\log (3x+15)=4 + \log 3$

Possible Answers:

41.416

17.021

21.544

16.544

Correct answer:

16.544

Explanation:

The expression can be simplified using logarithmic properties.

$\log(x^2+10x+25) +\log (3x+15)=4 + \log 3 \\$

$\log(x+5)^2 +\log (3*(x+5))=4 + \log 3$

$2\log(x+5) +\log 3 + \log(x+5)= 4 + \log 3$

$3 \log(x+5)=4$

We use the definition of common logarithm to solve for x.

(x+5)^3=10^4

$(x+5)=\sqrt[3]{10000}=21.544$

So x = 21.544 -5 or 16.544.

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Example Question #23 : Properties Of Logarithms

Solve for the value of x by simplifying the following logarithmic function:

log (x)+log(x+3) =1

Possible Answers:

x=3

x=2

x=-2

x=5

x=-5

Correct answer:

x=5

Explanation:

Log Property: Log(x) + Log(y) = Log(xy)

The equation log (x)+log(x+3) =1 can be simplified using the log property stated above.

The simplified version of the equation is:

log((x)(x-3))=1

Substitute Log(10) in for 1, since Log(10) = 1

log((x)(x-3))=log(10)

Next, remove the logs from both sides

(x)(x-3)=10

Solve for the value of x:

x(x-3)=10

$x^{2}-3x=10$

$x^{2}-3x-10=0$

(x-5)(x+2)=0

x=5, x=-2

A negative number cannot be evaluated in the log function equation above so the answer is x=5

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Example Question #24 : Properties Of Logarithms

Find the value of x to satisfy $log{_{16}}{4^{x}}=2$ .

Possible Answers:

x=16

x=12

x=2

x=4

x=3

Correct answer:

x=4

Explanation:

First, remove the log from the equation by changing the form:

$log{_{16}}{4^{x}}=2$

$16^{2}=4^{x}$

Next, change both sides so that they share a common base. In this case the common base is 4.

$(4^{2})^{2}=4^{x}$

Reduce the amount of exponents on the left side. Since an exponent is raised to an exponent you multiply the exponents together.

$4^{4}=4^{x}$

Set the exponents on both sides equal to eachother and solve for x.

4=x

Answer: x=4

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Example Question #25 : Properties Of Logarithms

Simplify

$x=\log_{16}\left(\frac{1}{64}\right)$ .

Possible Answers:

$x= \frac{-1}{4}$

x=4

$x= \frac{1}{4}$

x=-4

$x=-\frac{3}{2}$

Correct answer:

$x=-\frac{3}{2}$

Explanation:

Since the logarithmic function $y=\log_{b}(x)$ is the inverse of the exponential function x=b^y , we can make a straightforward simplification as follows:

$\frac{1}{64}=16^x$

64^-^1=16^x

$2^-^6=2^{4x}$

$-6 \log2=4x \log2$

-6=4x

$x=-\frac{3}{2}$

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Example Question #144 : Pre Calculus

Solve the equation,

$x=\log_{8}2+\log_{8}4+\log_{8}8$ .

Possible Answers:

Correct answer:

Explanation:

Using the logarithmic product rule $\log_b(xy)= \log_bx + \log_b y$ , we simplify as follows:

$\log_{8}2+\log_{8}4+\log_{8}8 = \log_8(8\cdot4\cdot2)$

$\log_864$

8^x=64

x=2

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Example Question #21 : Properties Of Logarithms

Simplify the expression $\sqrt{\log_4{16x}}$ .

Possible Answers:

$1\frac{1}{2}log_4 x$

$2+\frac{1}{2}log_4 x$

$1+\frac{1}{2}log x$

$2+\log_4 x$

$1+\frac{1}{2}log_4 x$

Correct answer:

$1+\frac{1}{2}log_4 x$

Explanation:

Utilizing the power rule of logs we reduce the original expression to:

$\frac{1}{2}{}\log_4{16x}$

From there we use the product property to obtain:

$\frac{1}{2}(\log_416+log_4 x)$

Since 4 to the second power is 16 this expression yields

$\frac{1}{2}(2+log_4 x)$ , then we multiply through by the one half to obtain our final answer $\left(1+\frac{1}{2}log_4 x\right)$ .

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Example Question #142 : Pre Calculus

$\left(log_a \left(\frac{b}{cd}\right)\right)^e$ is the simplified form of which of the following expressions?

Possible Answers:

e(log_a b-log_a c-log_a d)

e(log_a c-log_a b-log_a d)

e(log_a d-log_a b-log_a c)

(log_a b-log_a c-log_a d)^e

e(log_a d-log_a c-log_a b)

Correct answer:

e(log_a b-log_a c-log_a d)

Explanation:

First use the power property of logs to get:

$e log_a\frac{b}{cd}$ .

Then use the quotient property to get:

e (log_a b -log_acd) .

Then finally use the product property and simplify to yield the final answer, e (log_a b -log_ac-log_ad) .

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Example Question #145 : Pre Calculus

Simplify.

$\frac{8^49^3}{2^{10}3^4}$

Possible Answers:

$\frac{18}{5}$

$\frac{864}{240}$

$2^{2}3^2$

Correct answer:

Explanation:

$\frac{8^49^3}{2^{10}3^4}$

We observe that 8=2^3 and 9=3^2 and substitute to yeild

$\frac{(2^3)^4(3^2)^3}{2^{10}3^4}$ .

Now we use the fact that $(a^b)^c=a^{b*c}$ to simplify

$\frac{2^{12}3^6}{2^{10}3^4}$ .

Since $\frac{a^b}{a^c}=a^{b-c}$ simplifying again yeilds

2^23^2 which is just $4\cdot9$

This yeilds a final answer of 36

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Precalculus : Exponential and Logarithmic Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #31 : Exponential And Logarithmic Functions

Example Question #32 : Exponential And Logarithmic Functions

Example Question #32 : Exponential And Logarithmic Functions

Example Question #23 : Properties Of Logarithms

Example Question #24 : Properties Of Logarithms

Example Question #25 : Properties Of Logarithms

Example Question #144 : Pre Calculus

Example Question #21 : Properties Of Logarithms

Example Question #142 : Pre Calculus

Example Question #145 : Pre Calculus

All Precalculus Resources

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