Question

Solve the equation: ${2^{2x + 3}} - 65({2^x} - 1) - 57 = 0$

Answer 1

Hint: Rearranging the terms we get a quadratic equation in ${2^x}$. We can name it any variable, say $t$. Solving the equation we get the value of $t$. Re-substituting for $t$ we get the value of ${2^x}$and so we can find the value of $x$.

Formula used:
For any $a$ we have, ${a^x} \times {a^y} = {a^{x + y}}$
Product of two terms equal to zero implies either of the terms is zero.

Complete step-by-step answer:
Given the equation ${2^{2x + 3}} - 65({2^x} - 1) - 57 = 0$
We can open the brackets.
This gives, ${2^{2x + 3}} - 65 \times {2^x} + 65 - 57 = 0$
$ \Rightarrow {2^{2x + 3}} - 65 \times {2^x} + 8 = 0 - - - (i)$
We know that, ${a^x} \times {a^y} = {a^{x + y}}$
So we can write, ${2^{2x + 3}} = {2^{2x}} \times {2^3} = 8 \times {2^{2x}}$
Substituting this in $(i)$ we get,
$8 \times {2^{2x}} - 65 \times {2^x} + 8 = 0 - - - (ii)$
Now put ${2^x} = t$. So we get ${2^{2x}} = {t^2}$.
This gives,
$8{t^2} - 65t + 8 = 0 - - - (iii)$
Rearranging terms in equation $(iii)$ we get,
$8{t^2} - 64t - t + 8 = 0$
Taking $8t$ from first two terms and $ - 1$ from last two terms common outside we get,
$8t(t - 8) - (t - 8) = 0$
$ \Rightarrow (t - 8)(8t - 1) = 0$
Product of two terms equal to zero implies either of the terms is zero.
$ \Rightarrow (t - 8) = 0{\text{ or }}(8t - 1) = 0$
So we have two cases; $t = 8$ and $8t = 1 \Rightarrow t = \dfrac{1}{8}$
Substituting back for $t$ we get,
${2^x} = 8$ or ${2^x} = \dfrac{1}{8}$
This gives $x = 3$ or $x = - 3$, since ${2^3} = 8$ and ${2^{ - 3}} = \dfrac{1}{{{2^3}}}$.
So we get the solutions for the given equation; ${2^{2x + 3}} - 65({2^x} - 1) - 57 = 0$
The answers are $x = 3$ or $x = - 3$.

Additional information:
We have other similar results in algebra.
$\dfrac{{{a^x}}}{{{a^y}}} = {a^{x - y}}$
${({a^x})^y} = {a^{xy}}$


Note: The given equation looks complicated in first view. But by simple rearrangements we get a quadratic equation in the standard form and could solve easily. The substitution for ${2^x}$ is an important step. Do not forget to find the value of $x$ after getting the value of $t$.
The quadratic equation $8{t^2} - 65t + 8 = 0$ can also be solved in another way.
For a second degree equation of the form $a{x^2} + bx + c = 0$, we have the solution
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$