Solve for x 7=6+e^(3-2x)

$7 = 6 + e^{3 - 2 x}$

Rewrite the equation as $6 + e^{3 - 2 x} = 7$ .

$6 + e^{3 - 2 x} = 7$

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

Subtract $6$ from both sides of the equation.

$e^{3 - 2 x} = 7 - 6$

Subtract $6$ from $7$ .

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

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

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

Expand the left side.

Expand $\ln (e^{3 - 2 x})$ by moving $3 - 2 x$ outside the logarithm.

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

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

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

Multiply $3 - 2 x$ by $1$ .

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

Simplify the right side.

The natural logarithm of $1$ is $0$ .

$3 - 2 x = 0$

Subtract $3$ from both sides of the equation.

$- 2 x = - 3$

Divide each term in $- 2 x = - 3$ by $- 2$ and simplify.

Divide each term in $- 2 x = - 3$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{- 3}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{- 3}{- 2}$

Divide $x$ by $1$ .

$x = \frac{- 3}{- 2}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x = \frac{3}{2}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{3}{2}$

Decimal Form:

$x = 1.5$

Mixed Number Form:

$x = 1 \frac{1}{2}$