Let =(1)/(4)+(1)/(8)+(1)/(16)+ and =(1)/(3)+(1)/(9)+(1)/(27)+ .

Question

Let $\alpha =\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots \infty$ and

$\beta =\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots \infty$. Then the value of

$(0.2{)}^{{\log }_{\sqrt{5}}(\alpha )}+(0.04{)}^{{\log }_{5}(\beta )}$ is equal to :

Accepted Answer

We first find the values of $\alpha $ and $\beta $.

Given

$\alpha =\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$

This is an infinite geometric progression with

$a=\frac{1}{4},\, r=\frac{1}{2}$

So,

$\alpha =\frac{a}{1-r}=\frac{\frac{1}{4}}{1-\frac{1}{2}}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$

Now,

$\beta =\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots$

This is also an infinite geometric progression with

$a=\frac{1}{3},\, r=\frac{1}{3}$

Thus,

$\beta =\frac{a}{1-r}=\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{\frac{1}{3}}{\frac{2}{3}}=\frac{1}{2}$

So both are equal to $\frac{1}{2}$.

Now the expression becomes

$(0.2{)}^{{\log }_{\sqrt{5}}(\frac{1}{2})}+(0.04{)}^{{\log }_{5}(\frac{1}{2})}$

Convert decimals into fractions:

$0.2=\frac{1}{5},\, 0.04=\frac{1}{25}={5}^{-2}$

So,

${(\frac{1}{5})}^{{\log }_{\sqrt{5}}(\frac{1}{2})}+{(\frac{1}{25})}^{{\log }_{5}(\frac{1}{2})}$

That is,

${5}^{-{\log }_{\sqrt{5}}(\frac{1}{2})}+{5}^{-2{\log }_{5}(\frac{1}{2})}$

Now use the change of base idea:

${\log }_{\sqrt{5}}(\frac{1}{2})=\frac{{\log }_{5}(\frac{1}{2})}{{\log }_{5}(\sqrt{5})}$

Since

${\log }_{5}(\sqrt{5})={\log }_{5}({5}^{1/2})=\frac{1}{2}$

therefore

${\log }_{\sqrt{5}}(\frac{1}{2})=\frac{{\log }_{5}(\frac{1}{2})}{1/2}=2{\log }_{5}(\frac{1}{2})$

Hence first term:

${5}^{-{\log }_{\sqrt{5}}(1/2)}={5}^{-2{\log }_{5}(1/2)}$

So both terms are same:

${5}^{-2{\log }_{5}(1/2)}+{5}^{-2{\log }_{5}(1/2)}=2\cdot {5}^{-2{\log }_{5}(1/2)}$

Now use the identity

${a}^{{\log }_{a}x}=x$

Then

${5}^{-2{\log }_{5}(1/2)}={5}^{{\log }_{5}(1/2{)}^{-2}}$

So let us simplify properly:

$-2{\log }_{5}(\frac{1}{2})={\log }_{5}{(\frac{1}{2})}^{-2}?$

Instead, use

$k{\log }_{a}b={\log }_{a}({b}^{k})$

Thus,

$-2{\log }_{5}(\frac{1}{2})={\log }_{5}({(\frac{1}{2})}^{-2})={\log }_{5}(4)$

Therefore,

${5}^{-2{\log }_{5}(1/2)}={5}^{{\log }_{5}4}=4$

So the total value is

$2	imes 4=8$

Hence, the required value is

$8$

So the correct option is Option C.

Solutions · 1