Question

The value of ${{\log }_{3}}9+{{\log }_{5}}25+{{\log }_{2}}8$ is
\[\begin{align}
  & A.4 \\
 & B.5 \\
 & C.6 \\
 & D.7 \\
\end{align}\]

Answer 1

Hint: In this question, we are given a logarithmic function and we need to evaluate it. For this, we will separate the terms and then find values for each term. Then we will put them in the original expression and calculate our desired value. We will use the following properties of logarithm.
$\begin{align}
  & \left( i \right){{\log }_{b}}{{x}^{m}}=m{{\log }_{b}}x \\
 & \left( ii \right){{\log }_{b}}b=1 \\
\end{align}$

Complete step by step answer:
Here we are given the logarithmic function as ${{\log }_{3}}9+{{\log }_{5}}25+{{\log }_{2}}8\cdots \cdots \cdots \left( 1 \right)$.
We need to find its value. For this, let us separate the terms and find values of the terms individually. Finding the value of ${{\log }_{3}}9$.
We know that 9 can be written as $3\times 3$ which is equal to ${{3}^{2}}$. So our number becomes ${{\log }_{3}}9={{\log }_{3}}{{3}^{2}}$.
Now we know that ${{\log }_{b}}{{x}^{m}}=m{{\log }_{b}}x$.
Here b = 3, x = 3 and m = 2.
So we get ${{\log }_{3}}9=2{{\log }_{3}}3$.
We also know that ${{\log }_{b}}b=1$ so we get:
${{\log }_{3}}9=2\times 1\Rightarrow {{\log }_{3}}9=2\cdots \cdots \cdots \left( 2 \right)$.
Hence value of ${{\log }_{3}}9$ is 2.
Finding the value of ${{\log }_{5}}25$.
We know that 25 can be written as $5\times 5$ which is equal to ${{5}^{2}}$. So our number become,
${{\log }_{5}}25={{\log }_{5}}{{5}^{2}}$.
Now let us apply ${{\log }_{b}}{{x}^{m}}=m{{\log }_{b}}x$.
Here b = 5, x = 5, m = 2.
So we get ${{\log }_{5}}25=2{{\log }_{5}}5$.
Applying ${{\log }_{b}}b=1$ we get,
${{\log }_{5}}25=2\times 1\Rightarrow {{\log }_{5}}25=2\cdots \cdots \cdots \left( 3 \right)$.
Hence value of ${{\log }_{5}}25$ is 2.
Finding the value of ${{\log }_{2}}8$.
We know that 8 can be written as $2\times 2\times 2$ which is equal to ${{2}^{3}}$. So our number becomes,
${{\log }_{2}}8={{\log }_{2}}{{2}^{3}}$.
Now let us apply ${{\log }_{b}}{{x}^{m}}=m{{\log }_{b}}x$.
Here b = 2, x = 2, m = 3.
So we get ${{\log }_{2}}8=3{{\log }_{2}}2$.
Applying ${{\log }_{b}}b=1$ we get,
${{\log }_{2}}8=3\times 1\Rightarrow {{\log }_{2}}8=3\cdots \cdots \cdots \left( 4 \right)$.
Hence the value of ${{\log }_{2}}8$ is 3.
Putting the values of ${{\log }_{3}}9,{{\log }_{5}}25,{{\log }_{2}}8$ from (2), (3), (4) into the expression (1) we get,
${{\log }_{3}}9+{{\log }_{5}}25+{{\log }_{2}}8=2+2+3=7$.

So, the correct answer is “Option D”.

Note: Students should know all logarithmic properties before solving this sum. Students can get confused with the base and power of logarithmic. If we are given log x then it means that the base is 10 and if we are given ln x then it means that the base of log is e. Take care of signs.