# Find the value of ${\log _3}5 \times {\log _{25}}9.$

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Hint: We are going to use basic logarithmic formulae to solve the given problem.

Complete step-by-step answer:

Given logarithmic expression is ${\log _3}5 \times {\log _{25}}9.$

Using base change formula ${\log _b}a = \dfrac{{{{\log }_x}a}}{{{{\log }_x}b}}$ , we get

$ \Rightarrow {\log _3}5 \times {\log _{25}}9$

$$ = \dfrac{{\log 5}}{{\log 3}} \times \dfrac{{\log 9}}{{\log 25}}$$

$ = \dfrac{{\log 5}}{{\log 3}} \times \dfrac{{\log {3^2}}}{{\log {5^2}}}$

Using power rule of logarithms ${\log _x}{a^n} = n{\log _n}a$ , to simplify the above term,

$ = \dfrac{{\log 5}}{{\log 3}} \times \dfrac{{2\log 3}}{{2\log 5}} = \dfrac{1}{1}$

= 1

$$\therefore $$ The value of $${\log _3}5 \times {\log _{25}}9$$ = 1.

Note: Power rule of logarithm is ${\log _x}{a^n} = n{\log _n}a$

Base change formula in logarithms is ${\log _b}a = \dfrac{{{{\log }_x}a}}{{{{\log }_x}b}}$. If no base is written we have to consider that as a natural logarithm with base 10. Or we can consider all logarithms having the same base.

Complete step-by-step answer:

Given logarithmic expression is ${\log _3}5 \times {\log _{25}}9.$

Using base change formula ${\log _b}a = \dfrac{{{{\log }_x}a}}{{{{\log }_x}b}}$ , we get

$ \Rightarrow {\log _3}5 \times {\log _{25}}9$

$$ = \dfrac{{\log 5}}{{\log 3}} \times \dfrac{{\log 9}}{{\log 25}}$$

$ = \dfrac{{\log 5}}{{\log 3}} \times \dfrac{{\log {3^2}}}{{\log {5^2}}}$

Using power rule of logarithms ${\log _x}{a^n} = n{\log _n}a$ , to simplify the above term,

$ = \dfrac{{\log 5}}{{\log 3}} \times \dfrac{{2\log 3}}{{2\log 5}} = \dfrac{1}{1}$

= 1

$$\therefore $$ The value of $${\log _3}5 \times {\log _{25}}9$$ = 1.

Note: Power rule of logarithm is ${\log _x}{a^n} = n{\log _n}a$

Base change formula in logarithms is ${\log _b}a = \dfrac{{{{\log }_x}a}}{{{{\log }_x}b}}$. If no base is written we have to consider that as a natural logarithm with base 10. Or we can consider all logarithms having the same base.

Last updated date: 24th Sep 2023

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