Solve for x 4+3^(x+1)=8

$4 + 3^{x + 1} = 8$

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

Subtract $4$ from both sides of the equation.

$3^{x + 1} = 8 - 4$

Subtract $4$ from $8$ .

$3^{x + 1} = 4$

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

$\ln (3^{x + 1}) = \ln (4)$

Expand $\ln (3^{x + 1})$ by moving $x + 1$ outside the logarithm.

$(x + 1) \ln (3) = \ln (4)$

Simplify the left side.

Simplify $(x + 1) \ln (3)$ .

Apply the distributive property.

$x \ln (3) + 1 \ln (3) = \ln (4)$

Multiply $\ln (3)$ by $1$ .

$x \ln (3) + \ln (3) = \ln (4)$

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

$x \ln (3) + \ln (3) - \ln (4) = 0$

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$x \ln (3) + \ln (\frac{3}{4}) = 0$

Subtract $\ln (\frac{3}{4})$ from both sides of the equation.

$x \ln (3) = - \ln (\frac{3}{4})$

Divide each term in $x \ln (3) = - \ln (\frac{3}{4})$ by $\ln (3)$ and simplify.

Divide each term in $x \ln (3) = - \ln (\frac{3}{4})$ by $\ln (3)$ .

$\frac{x \ln (3)}{\ln (3)} = \frac{- \ln (\frac{3}{4})}{\ln (3)}$

Simplify the left side.

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

Cancel the common factor.

$\frac{x \ln (3)}{\ln (3)} = \frac{- \ln (\frac{3}{4})}{\ln (3)}$

Divide $x$ by $1$ .

$x = \frac{- \ln (\frac{3}{4})}{\ln (3)}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{\ln (\frac{3}{4})}{\ln (3)}$

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{\ln (\frac{3}{4})}{\ln (3)}$

Decimal Form:

$x = 0.26185950 \dots$