Question

What are the roots of the equation ${4^x} - 3 \times {2^{x + 2}} + 32 = 0$ ?
(A) $1,2$
(B) $3,4$
(C) $2,3$
(D) $1,3$

Answer 1

Hint:
Start with using the identities of exponents such as ${a^{m + n}} = {a^m} \times {a^n}$ and ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$ in the given equation ${4^x} - 3 \times {2^{x + 2}} + 32 = 0$ to transform it into a quadratic equation with a variable $'{2^x}'$ . Now use the Quadratic formula to find the roots of this equation. Express the obtained roots in the form of ${2^a}$ and compare it with ${2^x}$ to find the values of variable ‘x’.

Complete Step by Step Solution:
Here in this problem, we are given an equation ${4^x} - 3 \times {2^{x + 2}} + 32 = 0$ and we need to find the solutions to the unknown $'x'$ in the equation.
The equation ${4^x} - 3 \times {2^{x + 2}} + 32 = 0$ looks unusual for the method that we already know to solve the equation. But we can try to change this equation in a form that can be easily solved
As we know that exponential identities that:
$ \Rightarrow {a^{m + n}} = {a^m} \times {a^n}$ and ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$
Therefore by using these above properties we can write:
$ \Rightarrow {4^x} = {\left( {{2^2}} \right)^x} = {2^{2x}}$ and ${2^{x + 2}} = {2^x} \times {2^2} = 4 \times {2^x}$
Now let’s substitute these values on the given equation
$ \Rightarrow {4^x} - 3 \times {2^{x + 2}} + 32 = 0 \Rightarrow {2^{2x}} - 12 \times {2^x} + 32 = 0$
Here we have an equation ${2^{2x}} - 12 \times {2^x} + 32 = 0$, which can also be written in form of a single variable, i.e. ${2^x}$ as:
$ \Rightarrow {2^{2x}} - 12 \times {2^x} + 32 = 0 \Rightarrow {\left( {{2^x}} \right)^2} - 12 \times {2^x} + 32 = 0$
Now let us assume the variable ${2^x}$ as some $'m'$. After using ${2^x} = m$ in our equation, we get:
$ \Rightarrow {\left( {{2^x}} \right)^2} - 12 \times {2^x} + 32 = 0 \Rightarrow {m^2} - 12m + 32 = 0$
This becomes a quadratic equation in a variable $'m'$ and hence can be solved easily using the Quadratic formula. According to the Quadratic formula, for an equation of the form $a{x^2} + bx + c = 0$ , the roots or the values of the variable ‘x’ can be given by:
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \Rightarrow x = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}{\text{ and }}x = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
Therefore, the Quadratic formula can be used to solve the equation ${m^2} - 12m + 32 = 0$ if we compare the equation with $a{x^2} + bx + c = 0$ then we have:
$ \Rightarrow x = m,a = 1,b = - 12{\text{ and }}c = 32$
Thus,
$ \Rightarrow m = \dfrac{{12 \pm \sqrt {{{\left( { - 12} \right)}^2} - 4 \times 1 \times 32} }}{{2 \times 1}} = \dfrac{{12 \pm \sqrt {144 - 128} }}{2}$
On solving it further and solving the radical sign, we get:
$ \Rightarrow m = \dfrac{{12 \pm \sqrt {144 - 128} }}{2} = \dfrac{{12 \pm \sqrt {16} }}{2} = \dfrac{{12 \pm 4}}{2}$
This gives us two different values of the variable $'m'$ as:
$ \Rightarrow m = \dfrac{{12 + 4}}{2} = \dfrac{{16}}{2} = 8{\text{ or }}m = \dfrac{{12 - 4}}{2} = \dfrac{8}{2} = 4$
Therefore, from the above quadratic equation, we get $m = 8,4$
But we assumed that ${2^x} = m$. So by using it again we get:
$ \Rightarrow {2^x} = m = 4{\text{ and }}{2^x} = m = 8$
As we know that square of $2$ is $4$ and cube of $2$ is $8$ , or we can write it as:
$ \Rightarrow 4 = {2^2}{\text{ and }}8 = {2^3}$
Using the above square and root in the value of ${2^x}$, we get:
$ \Rightarrow {2^x} = 4 = {2^2}{\text{ and }}{2^x} = 8 = {2^3}$
Now by comparing both sides, we can see the equivalency in the base of RHS and LHS. Thus, the value of the variable $'x'$ can be written as:
$ \Rightarrow {2^x} = {2^2} \Rightarrow x = 2$
Similarly, ${2^x} = {2^3} \Rightarrow x = 3$
Thus, we obtain the values as $x = 2,3$

Hence, the option (C) is the correct answer.

Note:
In this question, the use of the concept of exponents and the transformation of the mother equation was a crucial part. An alternative approach can be to use the method of splitting the middle term in the quadratic equation to find the roots of the equation. That is the equation ${m^2} - 12m + 32 = 0$ can also be solved by splitting the middle term $' - 12m'$ as: ${m^2} - 12m + 32 = 0 \Rightarrow {m^2} - 8m - 4m + 32 = 0 \Rightarrow \left( {m - 8} \right)\left( {m - 4} \right) = 0$