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Question:

If log₂x.log$\frac{x}{64}$2 = log$\frac{x}{16}$2. Then x is:

Explanation:

log₂x.log$\frac{x}{64}$2 = log$\frac{x}{16}$2

∴ $\frac{\log x}{\log x}$ × $\frac{\log 2}{\log \left({\displaystyle \frac{x}{64}}\right)}$ = $\frac{\log 2}{\log \left({\displaystyle \frac{x}{16}}\right)}$

∴ $\frac{\log x\times \log \left({\displaystyle \frac{x}{16}}\right)}{\log {\displaystyle \frac{x}{64}}}$ = log 2

∴ (log x) × $\frac{(\log x-\log 16)}{(\log x-\log 64)}$ = log 2

Let log x = t

∴ $\frac{t(t-\log 16)}{t-\log 64}$ = log 2

∴ t² - t log 16 = t log 2 - log 2. log 64

∴ t² - 4t log 2 = t log 2 - 6(log 2)²

∴ t² - 5t log 2 + 6(log 2)² = 0

∴ (t - 2 log 2) (t - 3 log 2) = 0

∴ t = log 4 or t = log 8

∴ x = 4 or x = 8

Hence, option (b).