Question

How do you solve ${{e}^{2x}}-6{{e}^{x}}-16=0$?

Answer 1

To solve the given question, we should know how to find the roots of a quadratic equation. For a general quadratic equation $a{{x}^{2}}+bx+c=0$, using the quadratic formula method we can find the roots of the equation as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. We can find the roots by substituting the coefficients in this formula.

Complete step-by-step solution:
We are given the equation ${{e}^{2x}}-6{{e}^{x}}-16=0$, we have to solve this equation. The given equation can also be expressed as,
${{\left( {{e}^{x}} \right)}^{2}}-6{{e}^{x}}-16=0$
Substituting ${{e}^{x}}=t$ in the above equation, it can be expressed as
$\Rightarrow {{t}^{2}}-6t-16=0$
We get a quadratic equation in t. We know that for a general quadratic equation $a{{x}^{2}}+bx+c=0$, using the formula method we can find the roots of the equation as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
Substituting the value of coefficients, we get

& \Rightarrow t=\dfrac{-(-6)\pm \sqrt{{{\left( -6 \right)}^{2}}-4(1)(-16)}}{2(1)} \\\ & \Rightarrow t=\dfrac{6\pm \sqrt{100}}{2} \\\ & \Rightarrow t=\dfrac{6\pm 10}{2} \\\ \end{aligned}$$ $$\Rightarrow t=\dfrac{6+10}{2}=\dfrac{16}{2}$$ or $$\Rightarrow t=\dfrac{6-10}{2}=\dfrac{-4}{2}$$ $$\therefore t=8$$ or $$t=-2$$ Using the substitution, we can say that $${{e}^{x}}=t$$. Hence, we have $${{e}^{x}}=8$$ or $${{e}^{x}}=-2$$. From the first equation, $${{e}^{x}}=8$$. Taking $$\ln $$ of both sides, we get $$\ln {{e}^{x}}=\ln 8$$. Using the property of logarithm which states that, $$\ln {{a}^{b}}=b\ln a$$ in the left side of the above equation it can be expressed as, $$\Rightarrow x\ln {{e}^{{}}}=\ln 8$$ As the argument and base of $$\ln e$$ are the same, its value is one. $$\therefore {{x}^{{}}}=\ln 8$$ **From second value, $${{e}^{x}}=-2$$. But as $${{e}^{x}}$$ always gives positive values as its output this equation has no solution. The only solution is $${{x}^{{}}}=\ln 8$$.** **Note:** After using substitution in the equation, we must check if the term for which we used substitution can take the value we got or not. If it can not take the values, then we have to exclude it from the solution. For example, here we have to exclude $${{e}^{x}}=-2$$ from the solution.