Solve for x 4e^(2x)-7e^x=15

$4 e^{2 x} - 7 e^{x} = 15$

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

$4 {(e^{x})}^{2} - 7 e^{x} = 15$

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

$4 u^{2} - 7 u = 15$

Solve for $u$ .

Subtract $15$ from both sides of the equation.

$4 u^{2} - 7 u - 15 = 0$

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 $a \cdot c = 4 \cdot - 15 = - 60$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 u$ .

$4 u^{2} - 7 u - 15 = 0$

Rewrite $- 7$ as $5$ plus $- 12$

$4 u^{2} + (5 - 12) u - 15 = 0$

Apply the distributive property.

$4 u^{2} + 5 u - 12 u - 15 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(4 u^{2} + 5 u) - 12 u - 15 = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $4 u + 5$ .

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

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

$4 u + 5 = 0$

$u - 3 = 0$

Set $4 u + 5$ equal to $0$ and solve for $u$ .

Set $4 u + 5$ equal to $0$ .

$4 u + 5 = 0$

Solve $4 u + 5 = 0$ for $u$ .

Subtract $5$ from both sides of the equation.

$4 u = - 5$

Divide each term in $4 u = - 5$ by $4$ and simplify.

Divide each term in $4 u = - 5$ by $4$ .

$\frac{4 u}{4} = \frac{- 5}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 u}{4} = \frac{- 5}{4}$

Divide $u$ by $1$ .

$u = \frac{- 5}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$u = - \frac{5}{4}$

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

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

$u - 3 = 0$

Add $3$ to both sides of the equation.

$u = 3$

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

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

Substitute $- \frac{5}{4}$ for $u$ in $u = e^{x}$ .

$- \frac{5}{4} = e^{x}$

Solve $- \frac{5}{4} = e^{x}$ .

Rewrite the equation as $e^{x} = - \frac{5}{4}$ .

$e^{x} = - \frac{5}{4}$

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

$\ln (e^{x}) = \ln (- \frac{5}{4})$

The equation cannot be solved because $\ln (- \frac{5}{4})$ is undefined.

Undefined

There is no solution for $e^{x} = - \frac{5}{4}$

No solution

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

$3 = e^{x}$

Solve $3 = e^{x}$ .

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

$e^{x} = 3$

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

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

Expand the left side.

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

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

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

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

Multiply $x$ by $1$ .

$x = \ln (3)$

The result can be shown in multiple forms.

Exact Form:

$x = \ln (3)$

Decimal Form:

$x = 1.09861228 \dots$

Solve for x 4e^(2x)-7e^x=15

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