Solve for n 3.4e^(2-2n)-9=-4

$3.4 e^{2 - 2 n} - 9 = - 4$

Step 1

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

Add $9$ to both sides of the equation.

$3.4 e^{2 - 2 n} = - 4 + 9$

Add $- 4$ and $9$ .

$3.4 e^{2 - 2 n} = 5$

Step 2

Divide each term in $3.4 e^{2 - 2 n} = 5$ by $3.4$ and simplify.

Divide each term in $3.4 e^{2 - 2 n} = 5$ by $3.4$ .

$\frac{3.4 e^{2 - 2 n}}{3.4} = \frac{5}{3.4}$

Simplify the left side.

Cancel the common factor of $3.4$ .

Cancel the common factor.

$\frac{3.4 e^{2 - 2 n}}{3.4} = \frac{5}{3.4}$

Divide $e^{2 - 2 n}$ by $1$ .

$e^{2 - 2 n} = \frac{5}{3.4}$

Simplify the right side.

Divide $5$ by $3.4$ .

$e^{2 - 2 n} = 1.47058823$

Step 3

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

$\ln (e^{2 - 2 n}) = \ln (1.47058823)$

Step 4

Expand the left side.

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

$(2 - 2 n) \ln (e) = \ln (1.47058823)$

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

$(2 - 2 n) \cdot 1 = \ln (1.47058823)$

Multiply $2 - 2 n$ by $1$ .

$2 - 2 n = \ln (1.47058823)$

Step 5

Subtract $2$ from both sides of the equation.

$- 2 n = \ln (1.47058823) - 2$

Step 6

Divide each term in $- 2 n = \ln (1.47058823) - 2$ by $- 2$ and simplify.

Divide each term in $- 2 n = \ln (1.47058823) - 2$ by $- 2$ .

$\frac{- 2 n}{- 2} = \frac{\ln (1.47058823)}{- 2} + \frac{- 2}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 n}{- 2} = \frac{\ln (1.47058823)}{- 2} + \frac{- 2}{- 2}$

Divide $n$ by $1$ .

$n = \frac{\ln (1.47058823)}{- 2} + \frac{- 2}{- 2}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$n = - \frac{\ln (1.47058823)}{2} + \frac{- 2}{- 2}$

Divide $- 2$ by $- 2$ .

$n = - \frac{\ln (1.47058823)}{2} + 1$

Step 7

The result can be shown in multiple forms.

Exact Form:

$n = - \frac{\ln (1.47058823)}{2} + 1$

Decimal Form:

$n = 0.80716875 \dots$

Solve for n 3.4e^(2-2n)-9=-4

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