Logarithm 1

The given expression is $\log_{2.5} \left( \frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}... \right)$

Consider $\left (\dfrac{1}{3} + \dfrac{1}{3^2} + \dfrac{1}{3^3}+ ..... \right)$

It is an infinite G.P where $a=\dfrac{1}{3} , r = \dfrac{1}{3}$

Now formula for infinite terms of a G.P is $S = \dfrac{a}{1-r}.$

Plugging the respective values , we get $S = \dfrac{1}{2}$

Therefore

$\log_{2.5} \left( \dfrac{1}{3}+\dfrac{1}{3^{2}}+\dfrac{1}{3^{3}}... \right)= \log_{2.5}{\left(\dfrac{1}{2}\right)}=\log_{(0.4)^{-1}}{(2^{-1})}=\log_{0.4}{2}.$

Now $\alpha =\log_{0.4}{2}$

$(0.16)^{\alpha} = (0.16)^{\Large \log_{0.4}{2}}= (0.4)^{\Large2 \times \log_{0.4}{2}} = (0.4)^{\Large \log_{0.4}{2^2}}=(0.4)^{\Large \log_{0.4}{4}}$

$=\boxed{4}$

First consider the part inside the logarithm. Let x = 1/3 + 1/3^2 + 1/3^3 + ....... Using infinite GP sum formula, 1/3 + 1/3^2 + 1/3^3 + ..... = 1/3//1-1/3 =1/2.
Therefore, α = log2.5 (1/2). Using logarithmic property, (5/2)^α = 1/2
=> (2/5)^α = 2. Now, 0.16 = 16/100 = 4/25 = (2/5)^2
=>(0.16)^α = (2/5)2α = 2^2 = 4