Let α=1/4+1/8+1/16+…∞ and β=1/3+1/9+1/27+…∞. Then the value of (0.2)^(log_(√5)(α))+(0.04)^(log₅(β)) is equal to:

Solution

Maths sir · 3 steps

Sum both GPs first, then write 0.2, 0.04 and √5 as powers of 5 so the logs cancel.

1. Sum the GPs. For an infinite GP with |r|<1, the sum is a/(1-r). So α = (1/4)/(1-1/2) = 1/2 and β = (1/3)/(1-1/3) = 1/2.
2. Everything in base 5. Write 0.2 = 5⁻¹, 0.04 = 5⁻² and log_(√5) x = 2 log₅ x. Now both terms have base 5, and 5^(log₅ x) = x finishes each one.

   (0.2)^(log_(√5) 1/2) = 5^(-2 log₅ 1/2) = (1/2)⁻² = 4, (0.04)^(log₅ 1/2) = 5^(-2 log₅ 1/2) = 4

3. So. The value is 4 + 4 = 8, which is option (c).

The idea to keep. When the bases are 0.2, 0.04 and √5, rewrite them all as powers of 5. Then use a^(logₐ x) = x and the log disappears.

Answer from the NTA final answer key. Solution by PrepWiser.