Question

If $ x=\log _{5}1000,y=\log _{7}2058 $ then;
A. $ x > y $
B. $ x < y $
C. $ x=y $
D. $ {{x}^{2}}=2y $

Answer 1

Hint: To compare the value of x and y we will first simplify the ‘x’ and ‘y’ using the log identities, which will make us compare the value. And in the end we will figure out the value of ‘x’ and ‘y’ and compare them whether they are equal, or hold some inequality relation.

Complete step by step answer:
Moving ahead with the question in a stepwise manner;
Simplifying the x and y separately we will get;
 $ \begin{align}
  & x={{\log }_{5}}1000 \\
 & x={{\log }_{5}}{{\left( 10 \right)}^{3}} \\
\end{align} $
Using the identity, $ {{\log }_{a}}{{m}^{n}}=n{{\log }_{a}}m $
We will get;
\[\begin{align}
  & x=3{{\log }_{5}}10 \\
 & x=3\left( {{\log }_{5}}5\times 2 \right) \\
\end{align}\]
Using the identity $ {{\log }_{a}}m\times n={{\log }_{a}}m+{{\log }_{a}}n $ we will get;
\[\begin{align}
  & x=3\left( {{\log }_{5}}5+{{\log }_{5}}2 \right) \\
 & x=3{{\log }_{5}}5+3{{\log }_{5}}2 \\
 & x=3{{\log }_{5}}5+{{\log }_{5}}{{2}^{3}} \\
 & x=3{{\log }_{5}}5+{{\log }_{5}}8 \\
\end{align}\]
As we know, $ {{\log }_{a}}a=1 $ so $ {{\log }_{5}}5=1 $ . So we can write it as;
\[\begin{align}
  & x=3(1)+{{\log }_{5}}8 \\
 & x=3+{{\log }_{5}}8 \\
\end{align}\]---Equation (i)
Similarly solving for y;
 $ y={{\log }_{7}}2058 $
As we can write $ 2058=7\times 7\times 7\times 6 $ , so replace it in upper equation to simplify it;
\[y=y={{\log }_{7}}{{7}^{3}}\times 6\]
Using the identity $ {{\log }_{a}}m\times n={{\log }_{a}}m+{{\log }_{a}}n $ we will get;
 $ y={{\log }_{7}}{{7}^{3}}+{{\log }_{7}}6 $
By using the identity, $ {{\log }_{a}}{{m}^{n}}=n{{\log }_{a}}m $
 $ y=3{{\log }_{7}}7+{{\log }_{7}}6 $
As we know, $ {{\log }_{7}}7=1 $ so $ {{\log }_{7}}7=1 $ . So we can write it as;
 $ \begin{align}
  & y=3\left( 1 \right)+{{\log }_{7}}6 \\
 & y=3+{{\log }_{7}}6 \\
\end{align} $ ---Equation (ii)
So now comparing Equation (i) and (ii);
\[x=3+{{\log }_{5}}8\]
So as we know that if in $ {{\log }_{a}}m,m>a $ then the value of $ {{\log }_{a}}m $ will be always greater than 1 otherwise it will be less than 1 but greater than zero. So here in our case \[{{\log }_{5}}8\]is given, in which the upper value is greater than the base of log, so we can say that it will always be greater than 1. So when we will add three in it so we will always get value greater than 4 i.e. \[3+{{\log }_{5}}8 > 4\]so we can write that x will always be greater than 4 i.e.
 $ x > 4 $
Similarly seeing for the y, we have
 $ y=3+{{\log }_{7}}6 $
In this $ {{\log }_{7}}6 $ , for which as explained above that if in $ {{\log }_{a}}m,m > a $ then the value of $ {{\log }_{a}}m $ will be always greater than 1 otherwise less than 1 but greater than zero. So here in our case $ {{\log }_{7}}6 $ is given, in which upper value is less than base of log, so we can say that it will always be less than 1 but greater than zero, i.e. $ 0 <{{\log }_{7}}6 < 1 $ . So when we will add three in it so we will always get value greater than 3 but less than 4, i.e. $ 3 < 3+{{\log }_{7}}6 < 4 $ so we can write that ‘y’ will always be less than 4, i.e. $ 3 < y < 4 $
So from here we can say that
 $ x > 4 $ and $ 3 < y < 4 $
So we can conclude that x will always be greater than y i.e. $ x>y $

So, the correct answer is “Option A”.

Note: We can also go with the process of finding the value of $ {{\log }_{5}}8 $ and $ {{\log }_{7}}6 $ to find the exact value of ‘x’ and ‘y’ which will also give us same answer, but that will not worth it. Moreover if there exists a relation between ‘x’ and ‘y’ of some equate rather than inequality then they won’t go such complex type rather they will simplify and look much similar to each other.