
If $x = {\log _3}5,y = {\log _{17}}25$ then which of the following is correct?
$\left( a \right)$ x < y
$\left( b \right)$ x = y
$\left( c \right)$ x > y
$\left( d \right)$ None of these.
Answer
587.4k+ views
Hint: In this particular question use the logarithmic property i.e. ${\log _a}b = \dfrac{{\log b}}{{\log a}}$ and also use that log of positive integers values is always positive, so if a > b then log a > log b, so use these concepts to reach the solution of the question.
Complete step-by-step answer:
Given data:
$x = {\log _3}5,y = {\log _{17}}25$.
Now we have to find out the relation between x and y i.e. which is greater than the other.
Now as we know that according to the logarithmic property, ${\log _a}b = \dfrac{{\log b}}{{\log a}}$ so use this property in the given expressions we have,
$ \Rightarrow x = \dfrac{{\log 5}}{{\log 3}}$............. (1)
And
$y = \dfrac{{\log 25}}{{\log 17}} = \dfrac{{\log {5^2}}}{{\log 17}} = \dfrac{{2\log 5}}{{\log 17}}$..................... (2), $\left[ {\because \log {a^b} = b\log a} \right]$
Now divide equation (1) by equation (2) we have,
$ \Rightarrow \dfrac{x}{y} = \dfrac{{\dfrac{{\log 5}}{{\log 3}}}}{{\dfrac{{2\log 5}}{{\log 17}}}}$
Now simplify this we have,
$ \Rightarrow \dfrac{x}{y} = \dfrac{{\log 5}}{{\log 3}} \times \dfrac{{\log 17}}{{2\log 5}} = \dfrac{{\log 17}}{{2\log 3}}$
Now the above equation is also written as,
$ \Rightarrow \dfrac{x}{y} = \dfrac{{\log 17}}{{\log {3^2}}} = \dfrac{{\log 17}}{{\log 9}}$, $\left[ {\because \log {a^b} = b\log a} \right]$
Now as we know that 17 > 9, so log 17 > log 9.
$ \Rightarrow \dfrac{{\log 17}}{{\log 9}} > 1$
$ \Rightarrow \dfrac{x}{y} = \dfrac{{\log 17}}{{\log 9}} > 1$
$ \Rightarrow \dfrac{x}{y} > 1$
$ \Rightarrow x > y$
So this is the required answer.
So, the correct answer is “Option C”.
Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic logarithmic properties which is all stated above, so first simplify both of the given equations using the properties as above, then divide them and simplify again using the properties as above we will get the required answer.
Complete step-by-step answer:
Given data:
$x = {\log _3}5,y = {\log _{17}}25$.
Now we have to find out the relation between x and y i.e. which is greater than the other.
Now as we know that according to the logarithmic property, ${\log _a}b = \dfrac{{\log b}}{{\log a}}$ so use this property in the given expressions we have,
$ \Rightarrow x = \dfrac{{\log 5}}{{\log 3}}$............. (1)
And
$y = \dfrac{{\log 25}}{{\log 17}} = \dfrac{{\log {5^2}}}{{\log 17}} = \dfrac{{2\log 5}}{{\log 17}}$..................... (2), $\left[ {\because \log {a^b} = b\log a} \right]$
Now divide equation (1) by equation (2) we have,
$ \Rightarrow \dfrac{x}{y} = \dfrac{{\dfrac{{\log 5}}{{\log 3}}}}{{\dfrac{{2\log 5}}{{\log 17}}}}$
Now simplify this we have,
$ \Rightarrow \dfrac{x}{y} = \dfrac{{\log 5}}{{\log 3}} \times \dfrac{{\log 17}}{{2\log 5}} = \dfrac{{\log 17}}{{2\log 3}}$
Now the above equation is also written as,
$ \Rightarrow \dfrac{x}{y} = \dfrac{{\log 17}}{{\log {3^2}}} = \dfrac{{\log 17}}{{\log 9}}$, $\left[ {\because \log {a^b} = b\log a} \right]$
Now as we know that 17 > 9, so log 17 > log 9.
$ \Rightarrow \dfrac{{\log 17}}{{\log 9}} > 1$
$ \Rightarrow \dfrac{x}{y} = \dfrac{{\log 17}}{{\log 9}} > 1$
$ \Rightarrow \dfrac{x}{y} > 1$
$ \Rightarrow x > y$
So this is the required answer.
So, the correct answer is “Option C”.
Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic logarithmic properties which is all stated above, so first simplify both of the given equations using the properties as above, then divide them and simplify again using the properties as above we will get the required answer.
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