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Use logarithms to solve.

e^(2x)-e^(x)-56=0
Enter the exact answer (i.e. keep your answer in exponential or logarithmic form, you do not need to calculate its numeric value). Enclose arguments of functions in parentheses and include a multiplication sign between terms. For example,
c^(**)ln(h).

Use logarithms to solve. $\newline$ $e^{2 x}-e^{x}-56=0$ $\newline$ Enter the exact answer (i.e. keep your answer in exponential or logarithmic form, you do not need to calculate its numeric value). Enclose arguments of functions in parentheses and include a multiplication sign between terms. For example, $c^{*} \ln (h)$ .

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Product property of logarithms

Full solution

Q. Use logarithms to solve.

\newline

e^{2 x}-e^{x}-56=0

\newline

Enter the exact answer (i.e. keep your answer in exponential or logarithmic form, you do not need to calculate its numeric value). Enclose arguments of functions in parentheses and include a multiplication sign between terms. For example,

c^{*} \ln (h)

.

Substitute with new variable: Let's substitute $e^{x}$ with a new variable, say $y$ . So we have $y = e^{x}$ . This will simplify the equation.
Quadratic equation: Now, our equation becomes $y^2 - y - 56 = 0$ , which is a quadratic equation.
Factor the equation: We can factor this quadratic equation: $(y - 8)(y + 7) = 0$ .
Solve for $y$ : Setting each factor equal to zero gives us $y - 8 = 0$ or $y + 7 = 0$ .
Consider positive value: Solving for $y$ , we get $y = 8$ or $y = -7$ . But since $y = e^{x}$ and $e^{x}$ is always positive, $y$ cannot be $-7$ .
Substitute back: So we only consider $y = 8$ . Now we substitute back $e^{x}$ for $y$ , getting $e^{x} = 8$ .
Take natural logarithm: To solve for $x$ , we take the natural logarithm of both sides: $\ln(e^{x}) = \ln(8)$ .
Solve for $x$ : Using the property of logarithms that $\ln(e^{x}) = x$ , we have $x = \ln(8)$ .

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