Question

Given $\log x=m+n$ and $\log y=m-n$ , express the value of $\log \dfrac{10x}{{{y}^{2}}}$ in terms of m and n.

Answer 1

Hint: We are given the expression $\log \dfrac{10x}{{{y}^{2}}}$ and we need to express it in terms of m and n, where $\log x=m+n$ and $\log y=m-n$ . We have to use various properties of logarithms like $\log \dfrac{a}{b}=\log a-\log b$ , $\log ab=\log a+\log b$ and $\log {{a}^{b}}=b\log a$ to solve this problem. Using these, we first convert the expression to $\log 10x-\log {{y}^{2}}$ . After that, we write $\log 10x$ as $\log 10+\log x$ and $\log {{y}^{2}}$ as $2\log y$ . Having got the expressions $\operatorname{logx},logy$ , we substitute them according to the given and then get our final answer.

Complete step-by-step solution:
It is given that
$\log x=m+n....\left( i \right)$
$\log y=m-n....\left( ii \right)$
The expression that we are given to express in terms of m and n is $\log \dfrac{10x}{{{y}^{2}}}$ . Using the property of logarithms that $\log \dfrac{a}{b}=\log a-\log b$ , we get,
$\Rightarrow \log \dfrac{10x}{{{y}^{2}}}=\log 10x-\log {{y}^{2}}$
Again, using the property of logarithms that $\log ab=\log a+\log b$ , we get,
$\Rightarrow \log 10x=\log 10+\log x....\left( iii \right)$
Again, using the property of logarithms that $\log {{a}^{b}}=b\log a$ , we get,
$\Rightarrow \log {{y}^{2}}=2\log y....\left( iv \right)$
Now, using equation (i) in equation (iii), we get,
$\Rightarrow \log 10x=\log 10+m+n....\left( v \right)$
Using equation (ii) in equation (iv), we get,
$\Rightarrow \log {{y}^{2}}=2\left( m-n \right)....\left( vi \right)$
The given expression thus becomes, after using equations (v) and (vi),
$\Rightarrow \log \dfrac{10x}{{{y}^{2}}}=\left( \log 10+m+n \right)-2\left( m-n \right)$
This can be simplified by multiplying the $2$ inside the bracket and then opening up the brackets. After doing so, we get,
$\Rightarrow \log \dfrac{10x}{{{y}^{2}}}=\log 10+m+n-2m+2n$
Performing the additions and subtractions, we get,
$\Rightarrow \log \dfrac{10x}{{{y}^{2}}}=\log 10-m+3n$
Thus, we can conclude that the given expression can be written as $\log 10-m+3n$ in terms of m and n.

Note: Understanding the necessary approach of the problem is what’s needed the most. At first, simplifying the expression to the simplest form possible is required. Else, it creates confusion. Also, students sometimes make mistakes in the formula. They write $\log \left( a+b \right)=\log a+\log b$ and $\log \left( a-b \right)=\log a-\log b$ . These should be avoided.