Question

Find the value of $ a $ if $ {{4}^{a}}+{{4}^{a+1}}={{4}^{a+2}}-176 $ .
A. 2
B. 1
C. 3
D. 4

Answer 1

Hint: We first take all the variables on one side and keep the constant on the other sides. Then we apply the indices formulas to take $ {{4}^{a}} $ as common. We find the equation and apply the formula of $ {{x}^{m}}={{x}^{n}}\Rightarrow m=n $ . We find the value of the variable.

Complete step by step solution:
We simplify the given equation $ {{4}^{a}}+{{4}^{a+1}}={{4}^{a+2}}-176 $ by taking all the variables on one side and keeping the constant on the other sides.
The given variables are in the power form.
So,
 $ {{4}^{a}}+{{4}^{a+1}}={{4}^{a+2}}-176\\
\Rightarrow {{4}^{a}}+{{4}^{a+1}}-{{4}^{a+2}}=-176\;
 $ .
Let the numbers be $ {{x}^{m}} $ and $ {{x}^{n}} $ . We take multiplication of these numbers.
The indices get added. So, $ {{x}^{m+n}}={{x}^{m}}\times {{x}^{n}} $ . If we get $ {{x}^{m}}={{x}^{n}} $ , then we have $ m=n $ .
We get $ {{4}^{a+1}}=4\times {{4}^{a}},{{4}^{a+2}}={{4}^{2}}\times {{4}^{a}}=16\times {{4}^{a}} $ .
We take the $ {{4}^{a}} $ common from the terms.
 $
   {{4}^{a}}+{{4}^{a+1}}-{{4}^{a+2}}=-176 \\
  \Rightarrow {{4}^{a}}+4\times {{4}^{a}}-16\times {{4}^{a}}=-176 \\
  \Rightarrow {{4}^{a}}\left( 1+4-16 \right)=-176 \;
 $
We simplify the equation $ {{4}^{a}}\left( 1+4-16 \right)=-176 $ and get $ {{4}^{a}}\times \left( -11 \right)=-176 $ .
We are dividing both sides with $ -11 $ and get $ {{4}^{a}}=\dfrac{-176}{-11}=16 $ .
We now need to convert 16 in the form of $ {{4}^{a}} $ .
Therefore, $ {{4}^{a}}=16={{4}^{2}} $ . This gives $ a=2 $ . The correct option is A.
So, the correct answer is “Option A”.

Note: The addition and subtraction for exponents works for taking common terms out depending on the values of the indices.
For numbers $ {{a}^{m}} $ and $ {{a}^{n}} $ , we have $ {{a}^{m}}\pm {{a}^{n}}={{a}^{m}}\left( 1\pm {{a}^{n-m}} \right) $ .the relation is independent of the values of $ m $ and $ n $ .