Question 114

# $$\log_2 x.\log_{\frac{x}{64}}2 = \log_{\frac{x}{16}}2$$; then $$x = ?$$

Solution

$$\log_ba\ =\ \frac{1}{\log_ab}$$

$$\log_2 x.\log_{\frac{x}{64}}2$$ = $$\frac{\log_2x}{\log_2\left(\frac{x}{64}\right)}=\frac{\log_2x}{\log_2x-6}$$

Similarly,

$$\log_{\frac{x}{16}}2$$=$$\frac{1}{\log_2x-4}$$

Let $$\log_2x=t$$ => $$\frac{t}{t-6}=\frac{1}{t-4}$$ => $$t^2-5t+6=0$$ => $$\left(t-2\right)\left(t-3\right)=0$$ => t=2,3

.'. $$\log_2x=2\ or\ 3\ =>\ x=4\ or\ 8.$$

Hence, B is correct.