Question

The solution(s) of $15{(2)^{x + 1}} + 15{(2)^{2 - x}} = 135$ is/are
(This question has multiple correct options).
A) $2$
B) $ - 2$
C) $1$
D) $ - 1$

Answer 1

Hint: Here, first of all we will find the simplified form of the given equation by using the laws of power and exponents and get the required resultant value of “$x$”.

Complete step by step solution:
Given expression: $15{(2)^{x + 1}} + 15{(2)^{2 - x}} = 135$
Find the factor of the term on the right hand side of the equation. Place $135 = 15 \times 9$
$15{(2)^{x + 1}} + 15{(2)^{2 - x}} = 15(9)$
Common factors from all the terms in the above equation cancels each other.
${(2)^{x + 1}} + {(2)^{2 - x}} = (9)$
The above equation can be rewritten using the concepts of the law of power and exponent which states that the powers can be split keeping the base as common.
${(2)^x}{(2)^1} + {(2)^2}{(2)^{ - x}} = (9)$
Now, use the inverse of the power and exponent formula which states that ${x^{ - a}} = \dfrac{1}{{{x^a}}}$
${(2)^x}{(2)^1} + \dfrac{{{{(2)}^2}}}{{{{(2)}^x}}} = (9)$
Let us assume ${(2)^x} = A$ ….. (I)
and place in the above equation –
$2A + \dfrac{{{{(2)}^2}}}{A} = (9)$
Simplify the above equation in the form of the quadratic equation –
$2{A^2} + 4 = 9A$
Move all the terms on one side of the equation. When you move any term from one side to another then the sign of the terms also changes. Positive term becomes negative term.
$2{A^2} - 9A + 4 = 0$
Split the middle term in the above equation –
$2{A^2} - 8A - A + 4 = 0$
Make the pair of first two terms and the last terms
$\underline {2{A^2} - 8A} - \underline {A + 4} = 0$
Find the common factors in the above expression
$
  2A(A - 4) - 1(A - 4) = 0 \\
  (A - 4)(2A - 1) = 0 \\
 $
Now we get two values-
$A = 4{\text{ or A = }}\dfrac{1}{2}$
Place the above values in the equation (I)
$
  {2^x} = 4 \\
  {2^x} = {2^2} \\
 $
Comparing the powers, both the sides of the equation
$ \Rightarrow x = 2$ …. (II)
$
  {2^x} = \dfrac{1}{2} \\
  {2^x} = {2^{ - 1}} \\
 $
Comparing the powers, both the sides of the equation
$ \Rightarrow x = ( - 1)$ …. (III)
Therefore, $x=2, -1$, Hence, from the given multiple choices the options (A)) and (D) are the correct.

Note:
Be good in using the formulas for the power and exponent to solve. Don’t be confused in the multiplicative inverse and the additive inverse and its applications. Always remember that when bases are the same on both the sides of the equation then the powers are equal.