Find All Complex Solutions 2^(3x+1)=7

$2^{3 x + 1} = 7$

Step 1

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

$\ln (2^{3 x + 1}) = \ln (7)$

Step 2

Expand $\ln (2^{3 x + 1})$ by moving $3 x + 1$ outside the logarithm.

$(3 x + 1) \ln (2) = \ln (7)$

Step 3

Simplify the left side.

Simplify $(3 x + 1) \ln (2)$ .

Apply the distributive property.

$3 x \ln (2) + 1 \ln (2) = \ln (7)$

Multiply $\ln (2)$ by $1$ .

$3 x \ln (2) + \ln (2) = \ln (7)$

Step 4

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

$3 x \ln (2) + \ln (2) - \ln (7) = 0$

Step 5

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

$3 x \ln (2) + \ln (\frac{2}{7}) = 0$

Step 6

Subtract $\ln (\frac{2}{7})$ from both sides of the equation.

$3 x \ln (2) = - \ln (\frac{2}{7})$

Step 7

Divide each term in $3 x \ln (2) = - \ln (\frac{2}{7})$ by $3 \ln (2)$ and simplify.

Divide each term in $3 x \ln (2) = - \ln (\frac{2}{7})$ by $3 \ln (2)$ .

$\frac{3 x \ln (2)}{3 \ln (2)} = \frac{- \ln (\frac{2}{7})}{3 \ln (2)}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x \ln (2)}{3 \ln (2)} = \frac{- \ln (\frac{2}{7})}{3 \ln (2)}$

Rewrite the expression.

$\frac{x \ln (2)}{\ln (2)} = \frac{- \ln (\frac{2}{7})}{3 \ln (2)}$

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

Cancel the common factor.

$\frac{x \ln (2)}{\ln (2)} = \frac{- \ln (\frac{2}{7})}{3 \ln (2)}$

Divide $x$ by $1$ .

$x = \frac{- \ln (\frac{2}{7})}{3 \ln (2)}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{\ln (\frac{2}{7})}{3 \ln (2)}$