Question

How do you solve \[{{\text{4}}^{{\text{2x + 3}}}}{\text{ = 1}}\] ?

Answer 1

Hint: We are given two sides of the expression in two different forms. So first we will make them in the same form. By form I mean in base and power form. So 1 can be written as a number to the power 0. That on LHS can be written in powers of 2 first. Then since bases become the same we can equate their powers. And this will give the answer. That is the value of x.

Complete step-by-step answer:
Given that, \[{{\text{4}}^{{\text{2x + 3}}}}{\text{ = 1}}\]
4 can be written as a square of 2.
 \[ \Rightarrow {\left( {{2^2}} \right)^{2x + 3}}{\text{ = 1}}\]
Now using the laws of indices \[{\left( {{a^m}} \right)^n} = {a^{mn}}\] we can write
 \[ \Rightarrow {2^{2\left( {2x + 3} \right)}}{\text{ = 1}}\]
On multiplying we get,
 \[ \Rightarrow {2^{4x + 6}}{\text{ = 1}}\]
Now we can write 1 as \[{{\text{2}}^0}\]
 \[ \Rightarrow {2^{4x + 6}}{\text{ = }}{{\text{2}}^0}\]
There we go. Now since bases are same we can equate the powers,
 \[ \Rightarrow 4x + 6{\text{ = 0}}\]
Now let's find the value of x,
 \[ \Rightarrow 4x = - 6\]
Dividing both sides by 2 we get
 \[ \Rightarrow 2x = - 3\]
Thus x is equal to \[ \Rightarrow x = \dfrac{{ - 3}}{2}\]
This is the solution.
So, the correct answer is “$x = \dfrac{{ - 3}}{2}$”.

Note: Remember we cannot directly start solving the problem because both sides are not in the same form. 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.