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Algebra II : Exponents

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Example Questions

Example Question #2 : Using E

Solve for

$9^{2x-1}=27$

Possible Answers:

$\frac{5}{4}$

$\frac{4}{5}$

Correct answer:

$\frac{5}{4}$

Explanation:

Step 1: Achieve same bases

$9^{2x-1}=27$

$(3^{2})^{2x-1}=3^{3}$

$3^{4x-2}=3^{3}$

Step 2: Drop bases and set exponents equal to eachother

4x-2=3

Step 3: Solve for

4x=5

$x=\frac{5}{4}$

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Example Question #1 : Using E

Solve for

$\left(\frac{1}{3} \right)^{x-2}=9^{x-3}$

Possible Answers:

$\frac{8}{3}$

Correct answer:

$\frac{8}{3}$

Explanation:

Step 1: Achieve same bases

$\left(\frac{1}{3}\right)^{x-2}=9^{x-3}$

$(3^{-1})^{x-2}=(3^{2})^{x-3}$

$3^{-x+2}=3^{2x-6}$

Step 2: Drop bases, set exponenets equal to eachother

-x+2=2x-6

Step 3: Solve for

2=3x-6

8=3x

$x=\frac{8}{3}$

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Example Question #2 : Using E

Solve: e^l^n^(^2^x^+^3^)

Possible Answers:

The answer does not exist.

$\frac{1}{2x+3}$

2x+3

Correct answer:

2x+3

Explanation:

To solve e^l^n^(^2^x^+^3^) , it is necessary to know the property of .

e^l^n^(^x^)=x

Since and the terms cancel due to inverse operations, the answer is what's left of the term.

The answer is: 2x+3

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Example Question #8 : Using E

Simplify: ln(-2x) = ln(3x+3)

Possible Answers:

$\textup{There is no solution.}$

$\frac{3}{5}$

$-\frac{3}{7}$

$-\frac{3}{5}$

$\frac{3}{7}$

Correct answer:

$-\frac{3}{5}$

Explanation:

In order to eliminate the natural log on both side, we will need to raise both sides as a power with a base of . This will cancel out the natural logs.

$e^{ln(-2x)} =e^ {ln(3x+3)}$

The equation will become:

-2x=3x+3

Subtract on both sides.

-2x-3x=3x+3-3x

Simplify both sides.

-5x=3

Divide both sides by negative five.

$\frac{-5x}{-5}=\frac{3}{-5}$

The answer is: $-\frac{3}{5}$

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Example Question #9 : Using E

Simplify: ln(2x-3) = ln(x^2)

Possible Answers:

$1\pm i\sqrt2$

-3,1

$2\pm 2i\sqrt2$

$\textup{No solution.}$

-1,3

Correct answer:

$1\pm i\sqrt2$

Explanation:

In order to cancel the natural logs, we will need to use as a base and raise both raise both sides as the quantity of the power.

$e^{ln(2x-3)} =e^{ ln(x^2)}$

The equation becomes: 2x-3 =x^2

Subtract and add three on both sides.

2x-3 -[2x]+(3)=x^2-[2x]+(3)

The equation becomes: x^2-2x+3 =0

Use the quadratic equation to solve for the possible roots.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-(-2)\pm \sqrt{(-2)^2-4(1)(3)}}{2(1)}$

Simplify the quadratic equation.

$x=\frac{2\pm \sqrt{4-12}}{2} = 1\pm \frac{\sqrt{-8}}{2} = 1\pm \frac{2i\sqrt{2}}{2}$

The answers are: $1\pm i\sqrt2$

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Example Question #10 : Using E

Simplify: ln(3x+9) = 8

Possible Answers:

$\frac{1}{3}e^8-9$

8e-3

$\frac{e^8-3}{8}$

$\frac{1}{9}e^8-9$

$\frac{1}{3}e^8-3$

Correct answer:

$\frac{1}{3}e^8-3$

Explanation:

In order to get rid of the natural log, we will need to use the exponential term as a base and raise both sides as the powers using this base.

$e^{ln(3x+9)} =e^ 8$

The equation becomes:

3x+9=e^8

Subtract nine from both sides.

3x+9-9=e^8-9

3x=e^8-9

Divide by three on both sides.

$\frac{3x}{3}=\frac{e^8-9}{3}$

Simplify both sides.

The answer is: $\frac{1}{3}e^8-3$

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Example Question #191 : Exponents

Simplify: ln(4x-3)= 2

Possible Answers:

$\frac{2e^2+1}{3}$

$\frac{2e^2-1}{3}$

$\frac{e^2-3}{4}$

$\frac{e^2+3}{4}$

$\frac{e^{5}+1}{4}$

Correct answer:

$\frac{e^2+3}{4}$

Explanation:

In order to solve for the x-variable, we will need to raise both sides as powers of base , since the natural log has a default base of .

$e^{ln(4x-3)}= e^2$

The equation becomes:

4x-3 = e^2

Add three on both sides.

4x-3+3 = e^2+3

4x= e^2+3

Divide by four on both sides.

$\frac{4x}{4}= \frac{e^2+3}{4}$

The equation is: $x=\frac{e^2+3}{4}$

The answer is: $\frac{e^2+3}{4}$

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Example Question #11 : Using E

On the day of a child's birth, a sum of money is to be invested into a certificate of deposit (CD) that draws $6.2\%$ annual interest compounded continuously. The plan is for the value of the CD to be at least $\$20,000$ on the child's $18^{th}$ birthday.

If the amount of money invested is to be a multiple of $\$1,000$ , what is the minimum that should be invested initially, assuming that there are no further deposits or withdrawals?

Possible Answers:

$\$9,000$

$\$8,000$

$\$7,000$

$\$10,000$

$\$6,000$

Correct answer:

$\$7,000$

Explanation:

If we let be the initial amount invested and be the annual interest rate of the CD expressed as a decimal, then at the end of years, the amount of money that the CD will be worth can be determined by the formula

$A = Pe^{rt}$

Substitute A = 20,000 , r = 0.062 , t = 18 , and solve for .

$20,000 = Pe^{0.062 \cdot 18}$

$P = \frac{20,000 }{e^{0.062 \cdot 18}} \approx \frac{20,000 }{e^{1.116}} \approx \frac{20,000 }{3.0526} \approx 6,551$

The minimum principal to be invested initially is $6,551. However, since we are looking for the multiple of $1,000 that guarantees a minimum final balance of $20,000, we round up to the nearest such multiple, which is $7,000 - the correct response.

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Example Question #1 : Simple Exponents

Simplify the following expression

$3^{8}\times3^{10}$

Possible Answers:

$3^{2}$

$3^{80}$

$3^{18}$

$3^{-2}$

Correct answer:

$3^{18}$

Explanation:

$3^{8}\times3^{10}=3^{8+10}=3^{18}$

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Example Question #2 : Simple Exponents

Simplify the following expression

$\frac{x^{7}}{x^{3}}$

Possible Answers:

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

$x^{4}$

$x^{10}$

$x^{-4}$

Correct answer:

$x^{4}$

Explanation:

$\frac{x^{7}}{x^{3}}=x^{7-3}=x^{4}$

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Algebra II : Exponents

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #2 : Using E

Example Question #1 : Using E

Example Question #2 : Using E

Example Question #8 : Using E

Example Question #9 : Using E

Example Question #10 : Using E

Example Question #191 : Exponents

Example Question #11 : Using E

Example Question #1 : Simple Exponents

Example Question #2 : Simple Exponents

All Algebra II Resources

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