Question

How do you solve \[{e^{{x^2}}} = {e^{13x}}.\dfrac{1}{{{e^{40}}}}\] ?

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 using negative rule of indices we will get a quadratic equation.

Complete step by step solution:
Given, \[{e^{{x^2}}} = {e^{13x}}.\dfrac{1}{{{e^{40}}}}\] .
Using negative law of indices is
 \[{e^{{x^2}}} = {e^{13x}}.{e^{ - 40}}\]
 \[{e^{{x^2}}} = {e^{13x - 40}}\]
Apply logarithm on both sides of the equation we have,
 \[\log \left( {{e^{{x^2}}}} \right) = \log \left( {{e^{13x - 40}}} \right)\]
As we said earlier the inverse of the exponential functions are called logarithm functions.
 \[{x^2} = 13x - 40\]
Rearranging we have,
 \[{x^2} - 13x + 40 = 0\]
Using factorization method we can split the middle term we have,
 \[{x^2} - 8x - 5x + 40 = 0\]
 \[x(x - 8) - 5(x - 8) = 0\]
Now taking \[(x - 8)\] common we have,
 \[\left( {x - 8} \right)(x - 5) = 0\]
By zero product principle we have,
 \[\left( {x - 8} \right) = 0\] and \[(x - 5) = 0\]
 \[ \Rightarrow x = 8\] and \[x = 5\] . This is the required answer.
So, the correct answer is “ \[ x = 8\] and \[x = 5\] ”.

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 \[\log {x^a} = a\log x\] .
Note: 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 quadratic formulas to solve the equation of factorization method files. That is by using \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .