Question

How do you solve ${e^{{x^2} - 1}} = {e^{20}}$ ?

Answer 1

Hint : Here we need to solve for ‘x’. When one term is raised to the power of another term, the function is called an exponential function, for example $a = {x^y}$ . The inverse of the exponential functions are called logarithm functions that are logarithm and exponential cancels out so we apply logarithm to the given equation and we will get a quadratic equation. We can solve the obtained quadratic equation by factorization.

Complete step by step solution:
Given, ${e^{{x^2} - 1}} = {e^{20}}$ .
Apply logarithm on both sides of the equation we have,
$\log \left( {{e^{{x^2} - 1}}} \right) = \log \left( {{e^{80}}} \right)$
As we said earlier the inverse of the exponential functions are called logarithm functions.
${x^2} - 1 = 80$
Adding 1 on both sides of the equation,
${x^2} = 80 + 1$
${x^2} = 81$
Taking square roots on both sides of the equation,
$\sqrt {{x^2}} = \pm \sqrt {81}$
We know that square and square root will cancels out,
$x = \pm \sqrt {81}$
We know that 81 is a perfect square,
$\Rightarrow x = \pm 9$ .
Thus the solution of ${e^{{x^2} - 1}} = {e^{20}}$ is $9$ and $- 9$ .
So, the correct answer is “ $9$ and $- 9$ ”.

Note : To solve this kind of problem we need to remember the laws of logarithms. Product rule of logarithm that is the logarithm of the product is the sum of the logarithms of the factors. That is $\log (x.y) = \log (x) + \log (y)$ . Quotient rule of logarithm that is the logarithm of the ratio of two quantities is the logarithm of the numerator minus the logarithm of the denominator. that is $\log \left( {\dfrac{x}{y}} \right) = \log x - \log y$ . Power rule of logarithm that is the logarithm of an exponential number is the exponent times the logarithm of the base. That is $\log {x^a} = a\log x$ .
After simplification we have a quadratic equation, that is a polynomial of degree 2, hence we have two roots or two solutions. In above we have a simple quadratic equation hence we simplified it directly. We use factorization methods or quadratic formulas to solve the equation. That is by using $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .