Question

How do you solve the exponential equation \[{9^{2x}} = {27^{x - 1}}\] ?

Answer 1

Hint: The expression so given is having base and some power. In order to solve this type of problem we have to first make the base the same so that their powers can be put equal or can be compared. Here both 9 and 27 are powers of 3. One of them is square and the other is a cube of the same number. So they can be written in the form of 3 as base. This will help in comparing the power. After this we will be able to find the value of x.

Complete step-by-step answer:
Given that is an exponential expression
  \[{9^{2x}} = {27^{x - 1}}\]
But we know that \[9 = {3^2}\] and \[27 = {3^3}\]
So we can rewrite the above expression as,
  \[ \Rightarrow {\left( {{3^2}} \right)^{2x}} = {\left( {{3^3}} \right)^{x - 1}}\]
Now using the laws of indices as \[{\left( {{a^n}} \right)^m} = {a^{mn}}\] we can rewrite the powers as,
  \[ \Rightarrow {\left( 3 \right)^{2 \times 2x}} = {\left( 3 \right)^{3\left( {x - 1} \right)}}\]
On multiplying we get,
  \[ \Rightarrow {\left( 3 \right)^{4x}} = {\left( 3 \right)^{3x - 3}}\]
Now the bases of both sides are same then we can compare the powers as
  \[ \Rightarrow 4x = 3x - 3\]
On taking the variables on one side,
  \[ \Rightarrow 4x - 3x = - 3\]
Subtracting we get,
  \[ \Rightarrow x = - 3\]
This is our final answer.
So, the correct answer is “x = - 3”.

Note: Remember we cannot directly start solving the problem because both sides are not in the same form. By same form here we mean that until the bases are not the same we cannot compare the power directly. Indices and powers are having different laws. We can see addition, subtraction, multiplication and division also in powers. Index or power also means how many times that particular base number is to be multiplied. But if the power is zero then the answer is not zero it is always 1. And for the power 1 it is the same number once.