Solve by Factoring 5-3^x=-40

$5 - 3^{x} = - 40$

Step 1

Add $40$ to both sides of the equation.

$5 - 3^{x} + 40 = 0$

Step 2

Add $5$ and $40$ .

$45 - 3^{x} = 0$

Step 3

Subtract $45$ from both sides of the equation.

$- 3^{x} = - 45$

Step 4

Divide each term in $- 3^{x} = - 45$ by $- 1$ and simplify.

Divide each term in $- 3^{x} = - 45$ by $- 1$ .

$\frac{- 3^{x}}{- 1} = \frac{- 45}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{3^{x}}{1} = \frac{- 45}{- 1}$

Divide $3^{x}$ by $1$ .

$3^{x} = \frac{- 45}{- 1}$

Simplify the right side.

Divide $- 45$ by $- 1$ .

$3^{x} = 45$

Step 5

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

$\ln (3^{x}) = \ln (45)$

Step 6

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

$x \ln (3) = \ln (45)$

Step 7

Divide each term in $x \ln (3) = \ln (45)$ by $\ln (3)$ and simplify.

Divide each term in $x \ln (3) = \ln (45)$ by $\ln (3)$ .

$\frac{x \ln (3)}{\ln (3)} = \frac{\ln (45)}{\ln (3)}$

Simplify the left side.

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

Cancel the common factor.

$\frac{x \ln (3)}{\ln (3)} = \frac{\ln (45)}{\ln (3)}$

Divide $x$ by $1$ .

$x = \frac{\ln (45)}{\ln (3)}$