Solve for x 40+30e^(-0.2x)=50

$40 + 30 e^{- 0.2 x} = 50$

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

Subtract $40$ from both sides of the equation.

$30 e^{- 0.2 x} = 50 - 40$

Subtract $40$ from $50$ .

$30 e^{- 0.2 x} = 10$

Divide each term in $30 e^{- 0.2 x} = 10$ by $30$ and simplify.

Divide each term in $30 e^{- 0.2 x} = 10$ by $30$ .

$\frac{30 e^{- 0.2 x}}{30} = \frac{10}{30}$

Simplify the left side.

Cancel the common factor of $30$ .

Cancel the common factor.

$\frac{30 e^{- 0.2 x}}{30} = \frac{10}{30}$

Divide $e^{- 0.2 x}$ by $1$ .

$e^{- 0.2 x} = \frac{10}{30}$

Simplify the right side.

Cancel the common factor of $10$ and $30$ .

Factor $10$ out of $10$ .

$e^{- 0.2 x} = \frac{10 (1)}{30}$

Cancel the common factors.

Factor $10$ out of $30$ .

$e^{- 0.2 x} = \frac{10 \cdot 1}{10 \cdot 3}$

Cancel the common factor.

$e^{- 0.2 x} = \frac{10 \cdot 1}{10 \cdot 3}$

Rewrite the expression.

$e^{- 0.2 x} = \frac{1}{3}$

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

$\ln (e^{- 0.2 x}) = \ln (\frac{1}{3})$

Expand the left side.

Expand $\ln (e^{- 0.2 x})$ by moving $- 0.2 x$ outside the logarithm.

$- 0.2 x \ln (e) = \ln (\frac{1}{3})$

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

$- 0.2 x \cdot 1 = \ln (\frac{1}{3})$

Multiply $- 0.2$ by $1$ .

$- 0.2 x = \ln (\frac{1}{3})$

Divide each term in $- 0.2 x = \ln (\frac{1}{3})$ by $- 0.2$ and simplify.

Divide each term in $- 0.2 x = \ln (\frac{1}{3})$ by $- 0.2$ .

$\frac{- 0.2 x}{- 0.2} = \frac{\ln (\frac{1}{3})}{- 0.2}$

Simplify the left side.

Cancel the common factor of $- 0.2$ .

Cancel the common factor.

$\frac{- 0.2 x}{- 0.2} = \frac{\ln (\frac{1}{3})}{- 0.2}$

Divide $x$ by $1$ .

$x = \frac{\ln (\frac{1}{3})}{- 0.2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{\ln (\frac{1}{3})}{0.2}$

Replace $e$ with an approximation.

$x = - \frac{\log_{2.71828182} (\frac{1}{3})}{0.2}$

Divide $1$ by $3$ .

$x = - \frac{\log_{2.71828182} (0.\overline{3})}{0.2}$

Log base $2.71828182$ of $0.\overline{3}$ is approximately $- 1.09861228$ .

$x = - \frac{- 1.09861228}{0.2}$

Divide $- 1.09861228$ by $0.2$ .

$x = - - 5.49306144$

Multiply $- 1$ by $- 5.49306144$ .

$x = 5.49306144$

Solve for x 40+30e^(-0.2x)=50

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