Solve for x 3e^(x-3)-1=5

$3 e^{x - 3} - 1 = 5$

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

Add $1$ to both sides of the equation.

$3 e^{x - 3} = 5 + 1$

Add $5$ and $1$ .

$3 e^{x - 3} = 6$

Divide each term in $3 e^{x - 3} = 6$ by $3$ and simplify.

Divide each term in $3 e^{x - 3} = 6$ by $3$ .

$\frac{3 e^{x - 3}}{3} = \frac{6}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 e^{x - 3}}{3} = \frac{6}{3}$

Divide $e^{x - 3}$ by $1$ .

$e^{x - 3} = \frac{6}{3}$

Simplify the right side.

Divide $6$ by $3$ .

$e^{x - 3} = 2$

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

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

Expand the left side.

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

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

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

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

Multiply $x - 3$ by $1$ .

$x - 3 = \ln (2)$

Add $3$ to both sides of the equation.

$x = \ln (2) + 3$

The result can be shown in multiple forms.

Exact Form:

$x = \ln (2) + 3$

Decimal Form:

$x = 3.69314718 \dots$