Solve for x 3e^(3-x)=15

$3 e^{3 - x} = 15$

Step 1

Divide each term in $3 e^{3 - x} = 15$ by $3$ and simplify.

Divide each term in $3 e^{3 - x} = 15$ by $3$ .

$\frac{3 e^{3 - x}}{3} = \frac{15}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 e^{3 - x}}{3} = \frac{15}{3}$

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

$e^{3 - x} = \frac{15}{3}$

Simplify the right side.

Divide $15$ by $3$ .

$e^{3 - x} = 5$

Step 2

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

$\ln (e^{3 - x}) = \ln (5)$

Step 3

Expand the left side.

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

$(3 - x) \ln (e) = \ln (5)$

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

$(3 - x) \cdot 1 = \ln (5)$

Multiply $3 - x$ by $1$ .

$3 - x = \ln (5)$

Step 4

Subtract $3$ from both sides of the equation.

$- x = \ln (5) - 3$

Step 5

Divide each term in $- x = \ln (5) - 3$ by $- 1$ and simplify.

Divide each term in $- x = \ln (5) - 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{\ln (5)}{- 1} + \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{\ln (5)}{- 1} + \frac{- 3}{- 1}$

Divide $x$ by $1$ .

$x = \frac{\ln (5)}{- 1} + \frac{- 3}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{\ln (5)}{- 1}$ .

$x = - 1 \cdot \ln (5) + \frac{- 3}{- 1}$

Rewrite $- 1 \cdot \ln (5)$ as $- \ln (5)$ .

$x = - \ln (5) + \frac{- 3}{- 1}$

Divide $- 3$ by $- 1$ .

$x = - \ln (5) + 3$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = - \ln (5) + 3$

Decimal Form:

$x = 1.39056208 \dots$

Solve for x 3e^(3-x)=15

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