Question

If \[ \alpha=\frac14+\frac18+\frac1{16}+\cdots \text{ up to infinity} \] \[ \beta=\frac13+\frac19+\frac1{27}+\cdots \text{ up to infinity} \] Then value of \[ (0.2)^{\log_5\alpha}+(0.04)^{\log_5\beta} \] is

• A. \(\frac12\)
• B. \(8\)
• C. \(3\)
• D. \(\frac34\)

Hint:

Convert decimals like \(0.2\) or \(0.04\) into powers of \(5\) to simplify logarithmic expressions.

Answer 1

The Correct Option is B

Concept:

Use the sum of an infinite geometric progression

\[ S = \frac{a}{1-r}, \quad |r|<1 \]

and logarithmic exponent properties.

Step 1: Find \(\alpha\)

\[ \alpha = \frac14 + \frac18 + \frac1{16} + \cdots \]

Here

\[ a = \frac14, \quad r = \frac12 \]

\[ \alpha = \frac{\frac14}{1 - \frac12} \]

\[ \alpha = \frac12 \]

Step 2: Find \(\beta\)

\[ \beta = \frac13 + \frac19 + \frac1{27} + \cdots \]

Here

\[ a = \frac13, \quad r = \frac13 \]

\[ \beta = \frac{\frac13}{1 - \frac13} \]

\[ \beta = \frac12 \]

Step 3: Evaluate first term

\[ (0.2)^{\log_5 \alpha} \]

Since

\[ 0.2 = \frac15 \]

\[ \left(\frac15\right)^{\log_5 \frac12} \]

\[ = 5^{-\log_5 \frac12} \]

\[ = \left(\frac12\right)^{-1} \]

\[ = 2 \]

Step 4: Evaluate second term

\[ (0.04)^{\log_5 \beta} \]

\[ 0.04 = \frac1{25} = 5^{-2} \]

\[ (5^{-2})^{\log_5 \frac12} \]

\[ = 5^{-2\log_5 \frac12} \]

\[ = \left(\frac12\right)^{-2} \]

\[ = 4 \]

Step 5: Final value

\[ 2 + 4 = 6 \]