Question

Solve the linear equation ${4^{x - 1}} - {3.2^{x - 1}} + 2 = 0$

Answer 1

Hint: One can think of solving this question by using the concept of a quadratic equation. The questions which look very typical are generally very easy, just we have to analyze the question properly so that we can match it with any standard format question. Here in this question, we can easily do it by substituting a particular value with a variable.

Complete step-by-step solution:
${4^{x - 1}} - {3.2^{x - 1}} + 2 = 0$
Let,
${2^{x - 1}} = y$
We can also write,
${2^{2\left( {x - 1} \right)}} - {3.2^{x - 1}} + 2 = 0$
Now, put the value of ${2^{x - 1}}$
${y^2} - 3y + 2 = 0$
We can also write $3y = 2y + y$
${y^2} - \left( {2y + y} \right) + 2 = 0$
$\Rightarrow {y^2} - 2y - y + 2 = 0$
Taking common y and $ - 1$
$y\left( {y - 2} \right) - 1\left( {y - 2} \right) = 0$
$\Rightarrow \left( {y - 1} \right)\left( {y - 2} \right) = 0$
Now, there are two cases when $y = 1$ and $y = 2$
Case I: $y = 1$
Now, put the value of y
${2^{x - 1}} = {2^0}$
Here, base is equal in both L.H.S and R.H.S.
On comparing powers in L.H.S and R.H.S
$x - 1 = 0$
$\Rightarrow x = 1$
Case II: $y = 2$
Now, put the value of y
${2^{x - 1}} = 2$
We can also write $2 = {2^1}$
So,
${2^{x - 1}} = {2^1}$
Here bases are equal in both L.H.S and R.H.S
On comparing powers in L.H.S and R.H.S
$x - 1 = 1$
$\Rightarrow x = 2$
Hence, the values of x are $1\,and\,2.$

Note: A power is a product of multiplying a number by itself. Usually, a power is represented with a base number and an exponent. The base number tells what number is being multiplied. The exponent, a small number written above and to the right of the base number, tells how many times the base number is being multiplied. In ${5^2}$, the number $5$ is called the base, and the number $2$ is called the exponent. The exponent corresponds to the number of times the base is used as a factor.
If two powers have the same base then we can multiply the powers. When we multiply two powers we add their exponents.