Using e - Algebra II

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Question

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

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

Answer

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 , , , 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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Question

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

Answer

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.

2. Solve for

$x=Pe^{rt}$

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

You currently have $24,596 in your account.

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Question

Solve for

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

Answer

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

Step 3: Solve for

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Question

Solve for

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

Answer

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

Step 3: Solve for

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

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Question

Solve for

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

Answer

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

Step 3: Solve for

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

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Question

Solve for

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

Answer

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

Step 3: Solve for x

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

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Question

Solve for

$2^{-x}=4$

Answer

Step 1: Achieve same bases

$2^{-x}=4$

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

Step 2: Drop bases, set exponents equal to eachother

Step 3: Solve for

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Question

Solve:

Answer

To solve , it is necessary to know the property of .

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

The answer is:

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Question

Simplify:

Answer

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:

Subtract on both sides.

Simplify both sides.

Divide both sides by negative five.

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

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

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Question

Simplify:

Answer

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:

Subtract and add three on both sides.

The equation becomes:

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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Question

Simplify:

Answer

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:

Subtract nine from both sides.

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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Question

Simplify:

Answer

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:

Add three on both sides.

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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