Question

\(2^x=(0.2)^y=100\), then implies \(\frac{1}{x}-\frac{1}{y}=\)

• A. 1
• B. \(\frac{1}{2}\)
• C. \(\frac{1}{3}\)
• D. \(\log_{10} 2\)

Answer 1

The Correct Option is B

We are given that \[ 2^x = (0.2)^y = 100 \] Let's first solve for \(x\) and \(y\). \[ 2^x = 100 \implies x \log 2 = \log 100 \implies x = \frac{\log 100}{\log 2} = \frac{2 \log 10}{\log 2} = \frac{2}{\log 2} \] Similarly, \[ (0.2)^y = 100 \implies y \log 0.2 = \log 100 \implies y = \frac{\log 100}{\log 0.2} = \frac{2 \log 10}{\log 0.2} = \frac{2}{\log 0.2} \] Now, using the relationship \(\log 0.2 = -\log 5\), \[ y = \frac{2}{-\log 5} = \frac{2}{\log 5} \] Thus, \[ \frac{1}{x} - \frac{1}{y} = \frac{1}{\frac{2}{\log 2}} - \frac{1}{\frac{2}{\log 5}} = \frac{\log 2}{2} - \frac{\log 5}{2} = \frac{\log 2 - \log 5}{2} = \frac{\log \frac{2}{5}}{2} \] Using the value \(\frac{2}{5} = 0.4\), we get: \[ \frac{\log \frac{2}{5}}{2} = \frac{1}{2} \]

The correct option is (B): \(\frac{1}{2}\)