Let and . Then the value of is equal to: [2026]

Question

Let $α = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + . . . \infty$ and $β = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + . . . \infty$ . Then the value of ${(0.2)}^{\log_{\sqrt{5}} (α)} + {(0.04)}^{\log_{5} (β)}$ is equal to: [2026]

Smart Achievers · Accepted Answer

(3)

We have, $α = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + . . . \infty$

$α = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{4} \times \frac{2}{2} = \frac{1}{2}$

and $β = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{3} \times \frac{3}{2} = \frac{1}{2}$

Now, ${(0.2)}^{\log_{\sqrt{5}} (α)} {(0.04)}^{\log_{5} (β)}$

$= {(\frac{1}{5})}^{\log_{\sqrt{5}} (2^{- 1})} {+ (\frac{4}{100})}^{\log_{5} (\frac{1}{2})}$

$= 5^{2 \log_{5} (2)} {+ (\frac{1}{5^{2}})}^{\log_{5} (\frac{1}{2})}$

$= {4 + (\frac{1}{2})}^{\log_{5} (5^{- 2})} {= 4 + (\frac{1}{2})}^{- 2}$

$= 4 + 2^{2} = 4 + 4 = 8$

Solution

Q.
Let $α = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + . . . \infty$ and $β = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + . . . \infty$ . Then the value of ${(0.2)}^{\log_{\sqrt{5}} (α)} + {(0.04)}^{\log_{5} (β)}$ is equal to: [2026]

1 4

2 5

3 8

4 25

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Solution

Q. Let α=14+18+116+...∞ and β=13+19+127+...∞. Then the value of (0.2)log5(α)+(0.04)log5(β) is equal to: [2026]

1 4

2 5

3 8

4 25

Ans. (3) We have, α=14+18+116+...∞ α=141–12=1412=14×22=12 and β=131–13=1323=13×32=12 Now, (0.2)log5(α)(0.04)log5(β) =(15)log5(2–1)+(4100)log5(12) =52log5(2)+(152)log5(12) =4+(12)log5(5–2)=4+(12)–2 =4+22=4+4=8

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Q.
Let $α = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + . . . \infty$ and $β = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + . . . \infty$ . Then the value of ${(0.2)}^{\log_{\sqrt{5}} (α)} + {(0.04)}^{\log_{5} (β)}$ is equal to: [2026]