Question

# If ${{\log }_{2}}\left( {{x}^{2}}+1 \right)+{{\log }_{13}}\left( {{x}^{2}}+1 \right)={{\log }_{2}}\left( {{x}^{2}}+1 \right){{\log }_{13}}\left( {{x}^{2}}+1 \right)$ , $\left( x\ne 0 \right)$ , then ${{\log }_{7}}\left( {{x}^{2}}+24 \right)$ is equal to(a) 1(b) 2(c) 3(d) 4

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Hint: First, to proceed with this type of question, firstly we should solve the given condition to get the value of x and then use that value to get the required answer. Then, before we proceed, we must know that a log function with any base can be converted into base 10 which is actually represented by no base by dividing the log value with its log of the base value. Then, by finding the value of x and substituting it in the given expression we get the required answer.

In this question, we are supposed to find the converted value of ${{\log }_{7}}\left( {{x}^{2}}+24 \right)$ when condition is given as ${{\log }_{2}}\left( {{x}^{2}}+1 \right)+{{\log }_{13}}\left( {{x}^{2}}+1 \right)={{\log }_{2}}\left( {{x}^{2}}+1 \right){{\log }_{13}}\left( {{x}^{2}}+1 \right)$ .
So, to proceed with this type of question, firstly we should solve the given condition to get the value of x and then use that value to get the required answer.
But, before we proceed, we must know that log function with any base can be converted into base 10 which is actually represented by no base by dividing the log value with its log of the base value.
Sp, to understand it clearly, applying the same to the given condition as:
$\dfrac{\log \left( {{x}^{2}}+1 \right)}{\log 2}+\dfrac{\log \left( {{x}^{2}}+1 \right)}{\log 13}=\dfrac{\log \left( {{x}^{2}}+1 \right)}{\log 2}\times \dfrac{\log \left( {{x}^{2}}+1 \right)}{\log 13}$
Now, we can easily solve the above expression to get the value of x as:
\begin{align} & \dfrac{log13\times \log \left( {{x}^{2}}+1 \right)+log2\times \log \left( {{x}^{2}}+1 \right)}{\log 2\log 13}=\dfrac{\log \left( {{x}^{2}}+1 \right)}{\log 2}\times \dfrac{\log \left( {{x}^{2}}+1 \right)}{\log 13} \\ & \Rightarrow log13\times \log \left( {{x}^{2}}+1 \right)+log2\times \log \left( {{x}^{2}}+1 \right)=log\left( {{x}^{2}}+1 \right)\times \log \left( {{x}^{2}}+1 \right) \\ & \Rightarrow \log 13+\log 2=\log \left( {{x}^{2}}+1 \right) \\ & \Rightarrow \log \left( 26 \right)=\log \left( {{x}^{2}}+1 \right) \\ & \Rightarrow 26={{x}^{2}}+1 \\ & \Rightarrow {{x}^{2}}=25 \\ & \Rightarrow x=5 \\ \end{align}
So, we get the value of x as 5.
Now, by substituting the value of x as 5 in ${{\log }_{7}}\left( {{x}^{2}}+24 \right)$ , we get:
\begin{align} & {{\log }_{7}}\left( {{5}^{2}}+24 \right)={{\log }_{7}}\left( 25+24 \right) \\ & \Rightarrow {{\log }_{7}}49 \\ \end{align}
Now, by applying the same rule used above, we get:
${{\log }_{7}}49=\dfrac{\log \left( 49 \right)}{\log 7}$
Now, we know 49 can be written as the square of 7 and then by applying the property of the log function which says that the power comes in multiplication for log functions as:
\begin{align} & \dfrac{\log {{\left( 7 \right)}^{2}}}{\log 7}=\dfrac{2\log 7}{\log 7} \\ & \Rightarrow 2 \\ \end{align}
So, we get the value of ${{\log }_{7}}\left( {{x}^{2}}+24 \right)$ as 2.
So, the correct answer is “Option B”.

Note: Now, to solve these types of questions we need to know some of the basic conversions of the log functions so that these types of questions become easy for us and we don’t make any mistakes in them. So, the basic conversions are:
\begin{align} & \log \left( ab \right)=\log a+\log b \\ & \dfrac{\operatorname{loga}}{\log b}=\log a-\log b \\ \end{align}
From these, first conversion is used in the question to get the value of x.