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Algebra II : Using e

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Example Question #3311 : Algebra Ii

Twelve years ago, your grandma put money into a savings account for you that earns $7.5\%$ interest annually and is continuously compounded. How much money is currently in your account if she initially deposited $\$10,000$ and you have not taken any money out?

Possible Answers:

$8,103

$24,596

$81,030

$10,778

$21,170

Correct answer:

$24,596

Explanation:

1. Use $x=Pe^{rt}$ where is the current amount, is the interest rate, is the amount of time in years since the initial deposit, and is the amount initially deposited.

r=0.075

P=$10,000

t=12

2. Solve for

$x=Pe^{rt}$

$x=10,000e^{(0.075\cdot12)}$

x=24,596.03

You currently have $24,596 in your account.

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Example Question #2 : Using E

Solve for

$5^{x}=125^{3x-1}$

Possible Answers:

$\frac{3}{8}$

$-\frac{3}{8}$

Correct answer:

$\frac{3}{8}$

Explanation:

Step 1: Achieve same bases

$5^{x}=125^{3x-1}$

$5^{x}=(5^{3})^{3x-1}$

$5^{x}=5^{9x-3}$

Step 2: Drop bases, set exponents equal to eachother

x=9x-3

Step 3: Solve for x

-8x=-3

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

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Example Question #3 : Using E

Solve for

$2^{-x}=4$

Possible Answers:

Correct answer:

Explanation:

Step 1: Achieve same bases

$2^{-x}=4$

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

Step 2: Drop bases, set exponents equal to eachother

-x=2

Step 3: Solve for

x=-2

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Example Question #4 : Using E

Solve for

$4^{x}=2^{5x-3}$

Possible Answers:

$\frac{5}{3}$

Correct answer:

Explanation:

Step 1: Achieve same bases

$4^{x}=2^{5x-3}$

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

$2^{2x}=2^{5x-3}$

Step 2: Drop bases and set exponents equal to eachother

2x=5x-3

Step 3: Solve for

-3x=-3

x=1

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Example Question #5 : 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 #6 : 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 #7 : 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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Algebra II : Using e

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #3311 : Algebra Ii

Example Question #2 : Using E

Example Question #3 : Using E

Example Question #4 : Using E

Example Question #5 : Using E

Example Question #6 : Using E

Example Question #7 : Using E

Example Question #8 : Using E

Example Question #9 : Using E

Example Question #10 : Using E

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