1. If $$log_{25}{5}$$ = a and $$log_{25}{15} $$ = b, then the value of $$log_{25}{27}$$ is:
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By: anil on 05 May 2019 02.39 am
$$log_{25}{5}$$ = a => a=1/2 $$log_{25}{15}$$ = $$log_{25}{3}+log_{25}{5}$$ = b
$$frac{1}{2} log_{5}{3} + frac{1}{2}$$ = b $$log_{5}{3}$$ = 2(b - $$frac{1}{2})$$.............(i) $$log_{25}{27}$$ = $$frac{3}{2} log_{5}{3}$$.........(ii)
Replacing $$log_{5}{3}$$ = 2(b - $$frac{1}{2})$$ in (ii) we get $$log_{25}{27}$$ = 3(b - $$frac{1}{2}$$)
We can write -$$frac{1}{2}$$ as (- 1 + $$frac{1}{2}$$) or (-1 + a) So, $$log_{25}{27}$$ = 3(b + a - 1) Hence, option C is the correct answer.
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$$frac{1}{2} log_{5}{3} + frac{1}{2}$$ = b $$log_{5}{3}$$ = 2(b - $$frac{1}{2})$$.............(i) $$log_{25}{27}$$ = $$frac{3}{2} log_{5}{3}$$.........(ii)
Replacing $$log_{5}{3}$$ = 2(b - $$frac{1}{2})$$ in (ii) we get $$log_{25}{27}$$ = 3(b - $$frac{1}{2}$$)
We can write -$$frac{1}{2}$$ as (- 1 + $$frac{1}{2}$$) or (-1 + a) So, $$log_{25}{27}$$ = 3(b + a - 1) Hence, option C is the correct answer.