Evaluate e^(x-1)-5=5

$e^{x - 1} - 5 = 5$

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

Add $5$ to both sides of the equation.

$e^{x - 1} = 5 + 5$

Add $5$ and $5$ .

$e^{x - 1} = 10$

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

$\ln (e^{x - 1}) = \ln (10)$

Expand the left side.

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

$(x - 1) \ln (e) = \ln (10)$

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

$(x - 1) \cdot 1 = \ln (10)$

Multiply $x - 1$ by $1$ .

$x - 1 = \ln (10)$

Add $1$ to both sides of the equation.

$x = \ln (10) + 1$

The result can be shown in multiple forms.

Exact Form:

$x = \ln (10) + 1$

Decimal Form:

$x = 3.30258509 \dots$