How do you solve $ {6^x} = 72 $ ?

Hint: When one term is raised to the power of another term, the function is defined as an exponential function, for example $ a = {x^y} $ so the given equation is an exponential function. To solve such functions, we use the concept of logarithm functions, the logarithm functions are the inverse of the exponential functions, they are of the form $ y = {\log _x}a $ . Thus an exponential function like the one given in the question can be solved using the laws or properties of the logarithm.

Complete step-by-step answer:
We have to solve $ {6^x} = 72 $
Taking the natural log on both sides –
 $ \ln {6^x} = \ln 72 $
From the laws of the logarithm, we know that –
 $
  \log {x^n} = n\log x \\
   \Rightarrow \ln {6^x} = x\ln 6 \;
  $
Using this value in the above equation, we get –
 $
  x\ln 6 = \ln 72 \\
   \Rightarrow x = \dfrac{{\ln 72}}{{\ln 6}} \\
   \Rightarrow x = 2.38685281 \;
  $
But we take the values up to 3 decimal places, so we round off the above answer.
 $ \Rightarrow x = 2.387 $
Hence when $ {6^x} = 72 $ , $ x = 2.387 $
So, the correct answer is “ $ x = 2.387 $ ”.

Note: Usually, we solve such types of questions by factorizing that side of the equation which is not raised to some power. We write that number as a product of the number that is the base of the power x. For example, let $ {6^x} = 216 $ , we factorize 216 as – $ 216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 $ , now we write this number as a product of the base number – $ 216 = 6 \times 6 \times 6 = {6^3} $ , on comparing the left-hand side and the right-hand side of the example, we get x=3. But in the given question, when we factorize 72, we get – $ 72 = 2 \times 2 \times 2 \times 3 \times 3 = 6 \times 6 \times 2 = {6^2} \times 3 $ , as this number is not completely expressed as a power of 6, we cannot solve it by this method and have to use the logarithm function.