Using e

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Algebra II › Using e

Questions 1 - 10

1

Simplify:

$-\frac{3}{5}$

$\frac{3}{5}$

$\frac{3}{7}$

$-\frac{3}{7}$

$\textup{There is no solution.}$

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:

Subtract on both sides.

Simplify both sides.

Divide both sides by negative five.

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

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

2

Simplify:

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

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

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

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

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

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:

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

3

Solve for

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

$\frac{5}{3}$

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

Step 3: Solve for

4

Solve for

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

$\frac{5}{4}$

$\frac{4}{5}$

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

Step 3: Solve for

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

5

Simplify:

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

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

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

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

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:

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

6

Solve for

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

$\frac{8}{3}$

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

Step 3: Solve for

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

7

Simplify:

$1\pm i\sqrt2$

$2\pm 2i\sqrt2$

$\textup{No solution.}$

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:

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: $1\pm i\sqrt2$

8

Solve for

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

$\frac{3}{8}$

$-\frac{3}{8}$

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

Step 3: Solve for x

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

9

Solve:

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

The answer does not exist.

Explanation

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:

10

Solve for

$2^{-x}=4$

Explanation

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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