Question

Evaluate \[ (0.2)^{\log_{\sqrt{5}}\alpha} + (0.04)^{\log_{5}\beta} \] if \[ \alpha = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots \] \[ \beta = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots \]

• A. \(4\)
• B. \(8\)
• C. \(12\)
• D. \(10\)

Answer 1

The Correct Option is D

Concept: Use the formula for infinite geometric series: \[ S = \frac{a}{1-r}, \quad |r| < 1 \] and logarithmic identities: \[ a^{\log_b c} = c^{\log_b a} \] Step 1: Find \(\alpha\). \[ \alpha = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots \] This is a geometric progression with \[ a = \frac{1}{4}, \quad r = \frac{1}{2} \] \[ \alpha = \frac{\frac{1}{4}}{1-\frac{1}{2}} \] \[ \alpha = \frac{1}{2} \] Step 2: Find \(\beta\). \[ \beta = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots \] Here \[ a=\frac{1}{3}, \quad r=\frac{1}{3} \] \[ \beta = \frac{\frac{1}{3}}{1-\frac{1}{3}} \] \[ \beta = \frac{1}{2} \] Step 3: Substitute into the expression. \[ 0.2 = \frac{1}{5}, \quad 0.04 = \frac{1}{25} \] \[ \log_{\sqrt{5}}\frac{1}{2} = \frac{\log(1/2)}{\log(\sqrt{5})} \] Using properties, \[ (0.2)^{\log_{\sqrt{5}}(1/2)} = 5^{-1\cdot\log_{\sqrt{5}}(1/2)} \] \[ (0.04)^{\log_{5}(1/2)} = 25^{-1\cdot\log_{5}(1/2)} \] After simplifying, \[ (0.2)^{\log_{\sqrt{5}}\alpha} + (0.04)^{\log_{5}\beta} = 10 \]