Solve for x 12e^(2x)-5=17e^x

$12 e^{2 x} - 5 = 17 e^{x}$

Subtract $17 e^{x}$ from both sides of the equation.

$12 e^{2 x} - 5 - 17 e^{x} = 0$

Rewrite $e^{2 x}$ as exponentiation.

$12 {(e^{x})}^{2} - 5 - 17 e^{x} = 0$

Substitute $u$ for $e^{x}$ .

$12 u^{2} - 5 - 17 u = 0$

Move $- 5$ .

$12 u^{2} - 17 u - 5 = 0$

Solve for $u$ .

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 12 \cdot - 5 = - 60$ and whose sum is $b = - 17$ .

Factor $- 17$ out of $- 17 u$ .

$12 u^{2} - 17 u - 5 = 0$

Rewrite $- 17$ as $3$ plus $- 20$

$12 u^{2} + (3 - 20) u - 5 = 0$

Apply the distributive property.

$12 u^{2} + 3 u - 20 u - 5 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(12 u^{2} + 3 u) - 20 u - 5 = 0$

Factor out the greatest common factor (GCF) from each group.

$3 u (4 u + 1) - 5 (4 u + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 u + 1$ .

$(4 u + 1) (3 u - 5) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$4 u + 1 = 0$

$3 u - 5 = 0$

Set $4 u + 1$ equal to $0$ and solve for $u$ .

Set $4 u + 1$ equal to $0$ .

$4 u + 1 = 0$

Solve $4 u + 1 = 0$ for $u$ .

Subtract $1$ from both sides of the equation.

$4 u = - 1$

Divide each term in $4 u = - 1$ by $4$ and simplify.

Divide each term in $4 u = - 1$ by $4$ .

$\frac{4 u}{4} = \frac{- 1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 u}{4} = \frac{- 1}{4}$

Divide $u$ by $1$ .

$u = \frac{- 1}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$u = - \frac{1}{4}$

Set $3 u - 5$ equal to $0$ and solve for $u$ .

Set $3 u - 5$ equal to $0$ .

$3 u - 5 = 0$

Solve $3 u - 5 = 0$ for $u$ .

Add $5$ to both sides of the equation.

$3 u = 5$

Divide each term in $3 u = 5$ by $3$ and simplify.

Divide each term in $3 u = 5$ by $3$ .

$\frac{3 u}{3} = \frac{5}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 u}{3} = \frac{5}{3}$

Divide $u$ by $1$ .

$u = \frac{5}{3}$

The final solution is all the values that make $(4 u + 1) (3 u - 5) = 0$ true.

$u = - \frac{1}{4}, \frac{5}{3}$

Substitute $- \frac{1}{4}$ for $u$ in $u = e^{x}$ .

$- \frac{1}{4} = e^{x}$

Solve $- \frac{1}{4} = e^{x}$ .

Rewrite the equation as $e^{x} = - \frac{1}{4}$ .

$e^{x} = - \frac{1}{4}$

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

$\ln (e^{x}) = \ln (- \frac{1}{4})$

The equation cannot be solved because $\ln (- \frac{1}{4})$ is undefined.

Undefined

There is no solution for $e^{x} = - \frac{1}{4}$

No solution

Substitute $\frac{5}{3}$ for $u$ in $u = e^{x}$ .

$\frac{5}{3} = e^{x}$

Solve $\frac{5}{3} = e^{x}$ .

Rewrite the equation as $e^{x} = \frac{5}{3}$ .

$e^{x} = \frac{5}{3}$

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

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

Expand the left side.

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

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

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

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

Multiply $x$ by $1$ .

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 0.51082562 \dots$

Solve for x 12e^(2x)-5=17e^x

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