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Precalculus : Exponential and Logarithmic Functions

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Example Questions

Example Question #1 : Exponential Equations And Inequalities

Solve

$\frac{e^{x^2-2}}{e^{3x}}=e^{2x-6}$ .

Possible Answers:

x=1

x=1$ or $4

x=4

x=0$ or 1

x=7$ or $4

Correct answer:

x=1$ or $4

Explanation:

After using the division rule to simplify the left hand side you can take the natural log of both sides.

If you then combine like terms you get a quadratic equation which factors to,

(x-1)(x-4)=0 .

Setting each binomial equal to zero and solving for we get the solution to be x=1$ or $4 .

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Example Question #1 : Use Logarithms To Solve Exponential Equations And Inequalities

Solve for x: 3^x=243

Possible Answers:

x=3

x=-1

x=5

x=1

x=4

Correct answer:

x=4

Explanation:

$3^x=3^4\rightarrow x=4$

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Example Question #6 : Exponential Equations And Inequalities

Solve for x in the following equation: $3^{5x-3}=9^{2x}$

Possible Answers:

x=3

x=5

x=1

x=0

x=-1

Correct answer:

x=3

Explanation:

$3^{5x-3}=3^{(2)(2x)}\rightarrow 5x-3=4x\rightarrow x=3$

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Example Question #2 : Use Logarithms To Solve Exponential Equations And Inequalities

Solve for x using the rules of logarithms: $\log_{3}2x-\log_{3}(x-3)=1$

Possible Answers:

x=2

x=9

x=3

x=4

x=1

Correct answer:

x=9

Explanation:

$log_3 \frac{2x}{x-3}=1\rightarrow 3^{log_3 \frac{2x}{x-3}}=3^1\rightarrow \frac{2x}{x-3}=3\rightarrow x=9$

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Example Question #8 : Exponential Equations And Inequalities

Solve for x: $\log_{4}(3x-2)=2$

Possible Answers:

x=6

x=4

x=3

x=2

x=5

Correct answer:

x=6

Explanation:

$log_4 (3x-2)=2\rightarrow 3x-2=4^2\rightarrow 3x=18\rightarrow x=6$

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Example Question #3 : Use Logarithms To Solve Exponential Equations And Inequalities

Simplify the log expression: logx*log2

Possible Answers:

2log2x

Cannot be simplified any further

log2x

log(x)^2

2logx

Correct answer:

Cannot be simplified any further

Explanation:

The logarithmic expression is as simplified as can be.

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Example Question #1 : Application Problems

If you invest into a savings account which earns an interest rate per year, how much would it take for your deposit to double?

Possible Answers:

$\frac{\ln 2}{\ln(1+i)}$ $\text{years}$

$10 \text{ years}$

$\frac{\ln (1+i)}{2^n}$ $\text{years}$

$2 \text{ years}$

Correct answer:

$\frac{\ln 2}{\ln(1+i)}$ $\text{years}$

Explanation:

The equation of the value for this problem is

$\small R(1+i)^n=2R$ .

We can divide by R to get

$\small \small (1+i)^n=2\,$ .

We want to solve for n in this case, which is the amount of years. If we use the natural log on both sides and properties of logarithms, we get

$\small \ln(1+i)^n=n\ln(1+i)=\ln2$ .

If we solve for n, we get

$\small n = \frac{\ln(2)}{\ln(1+i)}$

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Example Question #1 : Application Problems

If you deposit $\$100$ into a savings account which earns a $5\%$ yearly interest rate, how much is in your account after two years?

Possible Answers:

$\$110$

$\$105$

$\$110.38$

$\$\small 110.25$

Correct answer:

$\$\small 110.25$

Explanation:

Since we are investing for two years with a yearly rate of 5%, we will use the formula to calculate compound interest.

A=P(1+r)^t

where

is the amount of money after time.

is the principal amount (initial amount).

is the interest rate.

is time.

Our amount after two years is:

$\small 100(1.05^2)=100(1.1025)=110.25$

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Example Question #3 : Application Problems

If you deposit into a savings account which compounds interest every month, what is the expression for the amount of money in your account after years if you earn a nominal interest rate of compounded monthly?

Possible Answers:

$\small R\bigg(1+\frac{i}{12}\bigg)^{24}$

$\small \small R\big(1+i\big)^{24}$

$\small \small R\big(1+i\big)^{2}$

$\small \small R\bigg(1+\frac{i}{12}\bigg)^{2}$

Correct answer:

$\small R\bigg(1+\frac{i}{12}\bigg)^{24}$

Explanation:

Since is the nominal interest rate compounded monthly we write the interest term as $\frac{i}{12}$ as it is the effective monthly rate.

We compound for years which is months. Since our interest rate is compounded monthly our time needs to be in the same units thus, months will be the units of time.

Plugging this into the equation for compound interest gives us the expression:

$A=P\left(1+\frac{i}{c}\right)^{ct}$

$A=R\left(1+\frac{i}{12}\right)^{12\cdot 2}$

$A=\small \small R \left(1+\frac{i}{12}\right)^{24}$

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Example Question #1 : Application Problems

John opens a savings account and deposits $\$1000$ into it. This savings account gains $5\%$ interest per year. After years, John withdraws all the money, and deposits it into another savings account with $6\%$ interest per year. years later, John withdraws the money.

How much money does John have after this year period? (Assume compound interest in both accounts)

Possible Answers:

$\$1118.12$

$\$1400.23$

$\$1276.28$

$\$1300.71$

$\$1157.63$

Correct answer:

$\$1300.71$

Explanation:

Plugging our numbers into the formula for compound interest, we have:

1000(1+.05)^3 = 1157.625 .

So John has about $\$1157.63$ after the first three years.

After placing his money into the other savings account, he has

1157.63(1+.06)^2 = 1300.713068 after more years.

So John has accumulated about $\$1300.71$ .

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Precalculus : Exponential and Logarithmic Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Exponential Equations And Inequalities

Example Question #1 : Use Logarithms To Solve Exponential Equations And Inequalities

Example Question #6 : Exponential Equations And Inequalities

Example Question #2 : Use Logarithms To Solve Exponential Equations And Inequalities

Example Question #8 : Exponential Equations And Inequalities

Example Question #3 : Use Logarithms To Solve Exponential Equations And Inequalities

Example Question #1 : Application Problems

Example Question #1 : Application Problems

Example Question #3 : Application Problems

Example Question #1 : Application Problems

All Precalculus Resources

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