Question

If \[\log \left( x+3 \right)-\log \left( x-3 \right)=1\] is $3\dfrac{m}{3}$, the value of m will be
\[\begin{align}
  & A.1 \\
 & B.2 \\
 & C.-1 \\
 & D.-2 \\
\end{align}\]

Answer 1

Hint: At first, solve the equation by using the formula,
\[\log c-\log d=\log \dfrac{c}{d}\]
Then, taking base value as 10. Then, finding the value of x and finally transform into a mixed fraction and then compare to find the value of m.

Complete step by step answer:
In the above question, we are given, the value of x is in the form of $3\dfrac{m}{3}$ such that it satisfies the given equation,
\[\log \left( x+3 \right)-\log \left( x-3 \right)=1\]
And hence, we have to find the value of m.
So, we are given the equation as,
\[\log \left( x+3 \right)-\log \left( x-3 \right)=1\]
We know an identity that,
\[\log x-\log y=\log \dfrac{x}{y}\]
So, instead of x, we will take x+3 and y we can take x-3, so we get,
\[\log \left( x+3 \right)-\log \left( x-3 \right)=1\]
As, \[\log \left( \dfrac{x+3}{x-3} \right)=1\]
Now, considering the base of log as 10, so we convert its value 1 as log 10, so we get,
\[\log \left( \dfrac{x+3}{x-3} \right)=\log 10\Rightarrow \dfrac{x+3}{x-3}=10\]
Now, on cross multiplication we get:
\[\begin{align}
  & x+3=10\left( x-3 \right) \\
 & \Rightarrow x+3=10x-30 \\
 & \Rightarrow 33=9x \\
\end{align}\]
Thus the value of x is $\dfrac{33}{9}\Rightarrow \dfrac{11}{3}$
We can convert improper to mixed fraction by writing it in form of $\text{Quotient}\dfrac{\text{Remainder}}{\text{Divisor}}$
So, $\dfrac{11}{3}$ can be written as $3\dfrac{2}{3}$
So, on comparing $3\dfrac{m}{3}$ with $3\dfrac{2}{3}$ we can say that value of m is 2.
Hence, the correct option is B.

Note:
Students generally confuse themselves while choosing base about what to choose as they should know that the default value of base is 10. So, they have to take it as 10 only, other values can make their answer wrong.