Question

Show that ${\log _3}5 > {\log _{17}}25$.

Answer 1

Hint: Here to find out which log function is greater, we will find the value of both log functions separately. For finding the value of ${\log _3}5$ and ${\log _{17}}25$, we will use the formula
$
 1) {\log _b}a = y \\
 2) a = {b^y} \\
 $

Complete step by step solution:
In this question, we are given an inequality between two log functions and we need to find out which one of them is greater than the other one.
Given inequality: ${\log _3}5 > {\log _{17}}25$
So, for finding out which one is greater from these two, we will find their log values separately.
So, first of all, let us find the value of ${\log _3}5$.
Now, we know the property of base that
$
   \Rightarrow {\log _b}a = y \\
   \Rightarrow a = {b^y} \\
 $
So, here first of all let ${\log _3}5$ equal to x. Therefore,
$
   \Rightarrow {\log _3}5 = x \\
   \Rightarrow 5 = {3^x} \\
 $
Now, to find the value of x, take log with base 10 on both sides. Therefore, we get
$ \Rightarrow {\log _{10}}5 = {\log _{10}}{3^x}$
Now, we know that $\log {a^b} = b\log a$. Therefore,
$
   \Rightarrow {\log _{10}}5 = x{\log _{10}}3 \\
   \Rightarrow 0.698970004 = x\left( {0.477121255} \right) \\
   \Rightarrow x = \dfrac{{0.698970004}}{{0.477121255}} \\
   \Rightarrow x = 1.464973 \\
 $
Hence, ${\log _3}5 = 1.464973$
Now, let us find the value of ${\log _{17}}25$.
Let the value of ${\log _{17}}25$ be y. Therefore,
$
   \Rightarrow {\log _{17}}25 = y \\
   \Rightarrow 25 = {17^y} \\
 $
Now, to find the value of y, take log with base 10 on both sides. Therefore, we get
$ \Rightarrow {\log _{10}}25 = {\log _{10}}{17^y}$
Now, we know that $\log {a^b} = b\log a$. Therefore,
$
   \Rightarrow {\log _{10}}25 = y{\log _{10}}17 \\
   \Rightarrow 1.397940009 = y\left( {1.230448921} \right) \\
   \Rightarrow y = \dfrac{{1.397940009}}{{1.230448921}} \\
   \Rightarrow y = 1.136121935 \\
 $
Hence, ${\log _{17}}25 = 1.136121935$.
Therefore, ${\log _3}5 > {\log _{17}}25$.

Note:
Another method to find the values of ${\log _3}5$ and ${\log _{17}}25$ is by changing the base to 10 by using the base change rule.
$ \Rightarrow {\log _c}x = \dfrac{{{{\log }_d}x}}{{{{\log }_d}c}}$
Therefore,
$ \Rightarrow {\log _3}5 = \dfrac{{{{\log }_{10}}5}}{{{{\log }_{10}}3}} = \dfrac{{0.698970004}}{{0.477121255}} = 1.4649$
$ \Rightarrow {\log _{17}}25 = \dfrac{{{{\log }_{10}}25}}{{{{\log }_{10}}17}} = \dfrac{{1.397940009}}{{1.230448921}} = 1.136121936$