If $$log_{25}{5}$$ = a and $$log_{25}{15} $$ = b, then the value of $$log_{25}{27}$$ is:
$$log_{25}{5}$$ = aÂ
=> 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 -Â $$\frac{1}{2})$$.............(i)
$$log_{25}{27}$$ = $$\frac{3}{2} \log_{5}{3}$$.........(ii)
Replacing $$\log_{5}{3}$$ = 2(b -Â $$\frac{1}{2})$$ in (ii) we get
$$log_{25}{27}$$ = 3(b -Â $$\frac{1}{2}$$)
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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