Find the value of ${\log _3}5 \times {\log _{25}}9.$
A. 0
B. 1
C. 2
D. 4
Last updated date: 19th Mar 2023
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Answer
305.7k+ views
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.
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