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

Solution

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

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

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

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

Therefore,Ā

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

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

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

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

=>Ā $$2^{2x}-20\cdot2^x+64=0$$

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

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

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

The values of $$2^$$ can'tĀ be 4 (log will be undefined), which implies The value of $$2^x$$ is 16.

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

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

The correct option is D

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