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# What is the value of ${{(a+b)}^{2}}$, If ${{5}^{a+b}}=5\times 25\times 125$ ?A.$25$B.$28$C.36D.$44$

Last updated date: 19th Mar 2023
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Hint: We have to find ${{(a+b)}^{2}}$ . For that, we have been given ${{5}^{a+b}}=5\times 25\times 125$ . So put $\log$ on both sides and use the properties of logarithm. After that, you will get the value of $(a+b)$ , then square it and you will get the answer.

Complete step-by-step Solution:

You must have come across the expression ${{3}^{2}}$. Here $3$ is the base and $2$ is the exponent. Exponents are also called Powers or Indices. The exponent of a number tells how many times to use the number in a multiplication. Let us study the laws of the exponent. It is very important to understand how the laws of exponents' laws are formulated.
Product law: According to the product law of exponents when multiplying two numbers that have the same base then we can add the exponents.
${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$
where a, m, and n all are natural numbers. Here the base should be the same in both the quantities.
Quotient Law: According to the quotient law of exponents, we can divide two numbers with the same base by subtracting the exponents. In order to divide two exponents that have the same base, subtract the power in the denominator from the power in the numerator.
$\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$
Power Law: According to the power law of exponents, if a number raises a power to a power, just multiply the exponents.
${{({{a}^{m}})}^{n}}={{a}^{mn}}$
Exponential form refers to a numeric form that involves exponents. One method to write such a number is by identifying that each position is representing a power (exponent) of $10$. Thus, you can initially break it up into different pieces. Exponents are also called Powers or Indices.
Now we have been given, ${{5}^{a+b}}=5\times 25\times 125$ .
So solving it,
${{5}^{a+b}}=5\times 25\times 125$
Applying $\log$on both sides we get,
$\log {{5}^{a+b}}=\log \left( 5\times 25\times 125 \right)$
We know the properties of $\log$ ,
$\log {{a}^{k}}=k\log a$
$\log (a\times b)=\log a+\log b$
$\log (\dfrac{a}{b})=\log a-\log b$
So now applying above properties we get,
\begin{align} & (a+b)\log 5=\log 5+\log 25+\log 125 \\ & (a+b)\log 5=\log 5+\log {{5}^{2}}+\log {{5}^{3}} \\ & (a+b)\log 5=\log 5+2\log 5+3\log 5 \\ \end{align}
Taking $\log 5$, in RHS we get,
\begin{align} & (a+b)\log 5=\log 5(1+2+3) \\ & (a+b)=1+2+3 \\ & (a+b)=6 \\ \end{align}
Now squaring both sides we get,
\begin{align} & {{(a+b)}^{2}}={{6}^{2}} \\ & {{(a+b)}^{2}}=36 \\ \end{align}
So we get the value of ${{(a+b)}^{2}}$ as $36$ .