The value of ( 0.16 )^(log _(2.5)( 1/3 + 1/3² + ....to ∞ )) is equal to ______.

Solution

Maths sir · 3 steps

Sum the GP first, then write 0.16 as a power of 2.5 so the base and the log cancel.

1. Sum the GP. The series has first term a = 1/3 and common ratio r = 1/3. Since |r| < 1, the infinite sum is a/(1-r) = (1/3)/(2/3) = 1/2.
2. Rewrite the base. 0.16 = 16/100 = (2/5)² and 2/5 = 1/2.5. So 0.16 = (2.5)⁻².

   (0.16)^(log_(2.5) 1/2) = (2.5)^(-2 log_(2.5) 1/2) = (2.5)^(log_(2.5)(1/2)⁻²) = (1/2)⁻²

3. So. Using a^(logₐ x) = x, the value is (1/2)⁻² = 2² = 4.

The common slip: 0.25. Writing 0.16 as (2.5)² instead of (2.5)⁻² gives (1/2)² = 0.25, not 4.

The idea to keep. When a log sits in an exponent, write the base as a power of the log's base. Then a^(logₐ x) = x makes the whole thing collapse.

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