Solve for x 40=3(2)^x-5

$40 = 3 {(2)}^{x} - 5$

Rewrite the equation as $3 \cdot 2^{x} - 5 = 40$ .

$3 \cdot 2^{x} - 5 = 40$

Move all terms not containing $x$ to the right side of the equation.

Add $5$ to both sides of the equation.

$3 \cdot 2^{x} = 40 + 5$

Add $40$ and $5$ .

$3 \cdot 2^{x} = 45$

Divide each term in $(3 \cdot 2^{x}) = 45$ by $3$ and simplify.

Divide each term in $(3 \cdot 2^{x}) = 45$ by $3$ .

$\frac{(3 \cdot 2^{x})}{3} = \frac{45}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 \cdot 2^{x}}{3} = \frac{45}{3}$

Divide $2^{x}$ by $1$ .

$2^{x} = \frac{45}{3}$

Simplify the right side.

Divide $45$ by $3$ .

$2^{x} = 15$

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

$\ln (2^{x}) = \ln (15)$

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

$x \ln (2) = \ln (15)$

Divide each term in $x \ln (2) = \ln (15)$ by $\ln (2)$ and simplify.

Divide each term in $x \ln (2) = \ln (15)$ by $\ln (2)$ .

$\frac{x \ln (2)}{\ln (2)} = \frac{\ln (15)}{\ln (2)}$

Simplify the left side.

Cancel the common factor of $\ln (2)$ .

Cancel the common factor.

$\frac{x \ln (2)}{\ln (2)} = \frac{\ln (15)}{\ln (2)}$

Divide $x$ by $1$ .

$x = \frac{\ln (15)}{\ln (2)}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (15)}{\ln (2)}$

Decimal Form:

$x = 3.90689059 \dots$