Question 48

For a real number x, if $$\frac{1}{2}, \frac{\log_3(2^x - 9)}{\log_3 4}$$, and $$\frac{\log_5\left(2^x + \frac{17}{2}\right)}{\log_5 4}$$ are in an arithmetic progression, then the common difference is


It is given that $$\frac{1}{2}, \frac{\log_3(2^x - 9)}{\log_3 4}$$, and $$\frac{\log_5\left(2^x + \frac{17}{2}\right)}{\log_5 4}$$ are in an arithmetic progression.

$$\frac{\log_3(2^x-9)}{\log_34}$$ can be written as $$\log_4\left(2^x-9\right)$$, and $$\frac{\log_5\left(2^x + \frac{17}{2}\right)}{\log_5 4}$$ can be written as $$\log_4\left(2^x+\frac{17}{2}\right)$$ 

Hence, $$2\log_4\left(2^x-9\right)=\frac{1}{2}+\log_4\left(2^x+\frac{17}{2}\right)$$

$$\frac{1}{2}$$ can be written as $$\log_42$$.


=> $$2\log_4\left(2^x-9\right)=\log_42+\log_4\left(2^x+\frac{17}{2}\right)$$

=> $$\log_4\left(2^x-9\right)^{^2}=\log_42\left(2^x+\frac{17}{2}\right)$$

=> $$\left(2^x-9\right)^{^2}=2\left(2^x+\frac{17}{2}\right)$$

=> $$2^{2x}-18\cdot2^x+81=2\cdot2^x+17$$

=> $$2^{2x}-20\cdot2^x+64=0$$

=> $$2^{2x}-16\cdot2^x-4\cdot2^x+64=0$$

=> $$2^x\left(2^x-16\right)-4\left(2^x-16\right)=0$$

=> $$\left(2^x-4\right)\left(2^x-16\right)=0$$

The values of $$2^$$ can't be 4 (log will be undefined), which implies The value of $$2^x$$ is 16.

Therefore, the common difference is $$\log_4\left(2^x-9\right)-\log_42$$

=> $$\log_47-\log_42\ =\ \log_4\left(\frac{7}{2}\right)$$

The correct option is D

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