Question

If we have an expression as ${{4}^{44}}+{{4}^{44}}+{{4}^{44}}+{{4}^{44}}={{4}^{x}},$ then x is __________.
$\begin{align}
  & A.\text{ 45} \\
 & \text{B}\text{. 44} \\
 & \text{C}\text{. 176} \\
 & \text{D}\text{. 11} \\
\end{align}$

Answer 1

Hint: In the given question it should be noted that ${{4}^{44}}$ added repeatedly, As, we know that repeated addition is multiplication so we have to write the above equation in multiplication form. So, we can write ${{4}^{44}}+{{4}^{44}}+{{4}^{44}}+{{4}^{44}}=4({{4}^{44}})$. As we know that when two same numbers are multiplied the powers of both numbers are added. In mathematical terms we can write
${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$. Also, when the bases of two equal numbers are equal, their powers also are equal. So, write the left-hand side and right-hand side terms, in the same base, once we write in the same base by equating its power, we get the solution of the given question.

Complete step-by-step answer:
It is given from question that
${{4}^{44}}+{{4}^{44}}+{{4}^{44}}+{{4}^{44}}={{4}^{x}}-----(1)$
As we know that repeated addition if multiplication so we can write
${{4}^{44}}+{{4}^{44}}+{{4}^{44}}+{{4}^{44}}=4({{4}^{44}})$
Now as we know from law of indices
${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$
So, we can write
$4({{4}^{44}})={{4}^{1+44}}={{4}^{45}}$
So, we can write from (1)
${{4}^{45}}={{4}^{x}}$
As we know from the law of indices when the bases of two equal numbers are equal, their powers also are equal.
Hence, we can write
$x=45$
So, option A is correct.

Note: It should be noted that when we use the formula of indices that is ${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$ the value of base that is $a$ cannot be equal to plus one or minus one. The value of $m$ and $n$ may be any real numbers.
In the step
${{4}^{45}}={{4}^{x}}$
We equate the power of 4 in order to get the solution, we can do it as another way by taking logarithm on both sides taking base as 4, as
${{\log }_{4}}{{4}^{45}}={{\log }_{4}}{{4}^{x}}$
Now as we know that ${{\log }_{a}}a=1\text{ and lo}{{\text{g}}_{a}}{{m}^{n}}=n{{\log }_{a}}m$
We can write
$\begin{align}
  & 45{{\log }_{4}}4=x{{\log }_{4}}4 \\
 & 45=x \\
\end{align}$
So, either by equating or by using logarithm we can get the solution.