Solve for x e^(3x-1)=2^x

$e^{3 x - 1} = 2^{x}$

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

$\ln (e^{3 x - 1}) = \ln (2^{x})$

Expand the left side.

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

$(3 x - 1) \ln (e) = \ln (2^{x})$

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

$(3 x - 1) \cdot 1 = \ln (2^{x})$

Multiply $3 x - 1$ by $1$ .

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

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

$3 x - 1 = x \ln (2)$

Subtract $x \ln (2)$ from both sides of the equation.

$3 x - 1 - x \ln (2) = 0$

Add $1$ to both sides of the equation.

$3 x - x \ln (2) = 1$

Factor the left side of the equation.

Factor $x$ out of $3 x - x \ln (2)$ .

Factor $x$ out of $3 x$ .

$x \cdot 3 - x \ln (2) = 1$

Factor $x$ out of $- x \ln (2)$ .

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

Factor $x$ out of $x \cdot 3 + x (- 1 \ln (2))$ .

$x (3 - 1 \ln (2)) = 1$

Rewrite $- 1 \ln (2)$ as $- \ln (2)$ .

$x (3 - \ln (2)) = 1$

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

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

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

Simplify the left side.

Cancel the common factor of $3 - \ln (2)$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 0.43349102 \dots$