Evaluate 7^(x-1)=e

$7^{x - 1} = e$

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

$\ln (7^{x - 1}) = \ln (e)$

Expand $\ln (7^{x - 1})$ by moving $x - 1$ outside the logarithm.

$(x - 1) \ln (7) = \ln (e)$

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

$(x - 1) \ln (7) = 1$

Simplify the left side.

Simplify $(x - 1) \ln (7)$ .

Apply the distributive property.

$x \ln (7) - 1 \ln (7) = 1$

Rewrite $- 1 \ln (7)$ as $- \ln (7)$ .

$x \ln (7) - \ln (7) = 1$

Add $\ln (7)$ to both sides of the equation.

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

Divide each term in $x \ln (7) = 1 + \ln (7)$ by $\ln (7)$ and simplify.

Divide each term in $x \ln (7) = 1 + \ln (7)$ by $\ln (7)$ .

$\frac{x \ln (7)}{\ln (7)} = \frac{1}{\ln (7)} + \frac{\ln (7)}{\ln (7)}$

Simplify the left side.

Cancel the common factor of $\ln (7)$ .

Cancel the common factor.

$\frac{x \ln (7)}{\ln (7)} = \frac{1}{\ln (7)} + \frac{\ln (7)}{\ln (7)}$

Divide $x$ by $1$ .

$x = \frac{1}{\ln (7)} + \frac{\ln (7)}{\ln (7)}$

Simplify the right side.

Cancel the common factor of $\ln (7)$ .

Cancel the common factor.

$x = \frac{1}{\ln (7)} + \frac{\ln (7)}{\ln (7)}$

Rewrite the expression.

$x = \frac{1}{\ln (7)} + 1$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{\ln (7)} + 1$

Decimal Form:

$x = 1.51389834 \dots$