Solve for x 3e^(x+1)-7=23

$3 e^{x + 1} - 7 = 23$

Step 1

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

Add $7$ to both sides of the equation.

$3 e^{x + 1} = 23 + 7$

Add $23$ and $7$ .

$3 e^{x + 1} = 30$

Step 2

Divide each term in $3 e^{x + 1} = 30$ by $3$ and simplify.

Divide each term in $3 e^{x + 1} = 30$ by $3$ .

$\frac{3 e^{x + 1}}{3} = \frac{30}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 e^{x + 1}}{3} = \frac{30}{3}$

Divide $e^{x + 1}$ by $1$ .

$e^{x + 1} = \frac{30}{3}$

Simplify the right side.

Divide $30$ by $3$ .

$e^{x + 1} = 10$

Step 3

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

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

Step 4

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 $1$ .

$x + 1 = \ln (10)$

Step 5

Subtract $1$ from both sides of the equation.

$x = \ln (10) - 1$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \ln (10) - 1$

Decimal Form:

$x = 1.30258509 \dots$