Precalculus Examples

$e^{n} = 3^{n + 2}$

Step 1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{n}) = \ln (3^{n + 2})$

Step 2

Expand the left side.

Tap for more steps...

Step 2.1

Expand $\ln (e^{n})$ by moving $n$ outside the logarithm.

$n \ln (e) = \ln (3^{n + 2})$

Step 2.2

The natural logarithm of $e$ is $1$ .

$n \cdot 1 = \ln (3^{n + 2})$

Step 2.3

Multiply $n$ by $1$ .

$n = \ln (3^{n + 2})$

Step 3

Expand $\ln (3^{n + 2})$ by moving $n + 2$ outside the logarithm.

$n = (n + 2) \ln (3)$

Step 4

Simplify the right side.

Tap for more steps...

Step 4.1

Apply the distributive property.

$n = n \ln (3) + 2 \ln (3)$

Step 5

Move all the terms containing a logarithm to the left side of the equation.

$- n \ln (3) - 2 \ln (3) = - n$

Step 6

Add $n$ to both sides of the equation.

$- n \ln (3) - 2 \ln (3) + n = 0$

Step 7

Add $2 \ln (3)$ to both sides of the equation.

$- n \ln (3) + n = 2 \ln (3)$

Step 8

Factor $n$ out of $- n \ln (3) + n$ .

Tap for more steps...

Step 8.1

Factor $n$ out of $- n \ln (3)$ .

$n (- 1 \ln (3)) + n = 2 \ln (3)$

Step 8.2

Raise $n$ to the power of $1$ .

$n (- 1 \ln (3)) + n = 2 \ln (3)$

Step 8.3

Factor $n$ out of $n^{1}$ .

$n (- 1 \ln (3)) + n \cdot 1 = 2 \ln (3)$

Step 8.4

Factor $n$ out of $n (- 1 \ln (3)) + n \cdot 1$ .

$n (- 1 \ln (3) + 1) = 2 \ln (3)$

Step 9

Rewrite $- 1 \ln (3)$ as $- \ln (3)$ .

$n (- \ln (3) + 1) = 2 \ln (3)$

Step 10

Divide each term in $n (- \ln (3) + 1) = 2 \ln (3)$ by $- \ln (3) + 1$ and simplify.

Tap for more steps...

Step 10.1

Divide each term in $n (- \ln (3) + 1) = 2 \ln (3)$ by $- \ln (3) + 1$ .

$\frac{n (- \ln (3) + 1)}{- \ln (3) + 1} = \frac{2 \ln (3)}{- \ln (3) + 1}$

Step 10.2

Simplify the left side.

Tap for more steps...

Step 10.2.1

Cancel the common factor of $- \ln (3) + 1$ .

Tap for more steps...

Step 10.2.1.1

Cancel the common factor.

$\frac{n (- \ln (3) + 1)}{- \ln (3) + 1} = \frac{2 \ln (3)}{- \ln (3) + 1}$

Step 10.2.1.2

Divide $n$ by $1$ .

$n = \frac{2 \ln (3)}{- \ln (3) + 1}$

Step 10.3

Simplify the right side.

Tap for more steps...

Step 10.3.1

Factor $- 1$ out of $- \ln (3)$ .

$n = \frac{2 \ln (3)}{- \ln (3) + 1}$

Step 10.3.2

Rewrite $1$ as $- 1 (- 1)$ .

$n = \frac{2 \ln (3)}{- \ln (3) - 1 (- 1)}$

Step 10.3.3

Factor $- 1$ out of $- \ln (3) - 1 (- 1)$ .

$n = \frac{2 \ln (3)}{- (\ln (3) - 1)}$

Step 10.3.4

Rewrite negatives.

Tap for more steps...

Step 10.3.4.1

Rewrite $- (\ln (3) - 1)$ as $- 1 (\ln (3) - 1)$ .

$n = \frac{2 \ln (3)}{- 1 (\ln (3) - 1)}$

Step 10.3.4.2

Move the negative in front of the fraction.

$n = - \frac{2 \ln (3)}{\ln (3) - 1}$

Step 11

The result can be shown in multiple forms.

Exact Form:

$n = - \frac{2 \ln (3)}{\ln (3) - 1}$

Decimal Form:

$n = - 22.28144795 \dots$