CAT 2023 Slot 3 — QA Question 4

Collapse to base 4. Using $\dfrac{\log_b P}{\log_b 4}=\log_4 P$, the three terms become

$$\frac12,\qquad \log_4(2^x-9),\qquad \log_4\!\left(2^x+\tfrac{17}{2}\right)$$

Apply the AP condition (twice the middle = sum of the ends):

$$\begin{aligned} &2\log_4(2^x-9)=\frac12+\log_4\!\left(2^x+\tfrac{17}{2}\right)\\ &\Rightarrow\ \log_4(2^x-9)^2=\log_4 2+\log_4\!\left(2^x+\tfrac{17}{2}\right)\\ &\Rightarrow\ (2^x-9)^2=2\left(2^x+\tfrac{17}{2}\right) \end{aligned}$$

(using $\tfrac12=\log_4 2$.)

Solve for $a=2^x$:

$$\begin{aligned} &(a-9)^2=2a+17\\ &\Rightarrow\ a^2-18a+81=2a+17\\ &\Rightarrow\ a^2-20a+64=0\\ &\Rightarrow\ (a-16)(a-4)=0\ \Rightarrow\ a=16 \text{ or } 4 \end{aligned}$$

Reject $a=4$ (it makes $2^x-9=-5\lt 0$, no log). So $2^x=16$.

Common difference = 2nd term $-$ 1st term, with $2^x-9=16-9=7$:

$$d=\log_4 7-\frac12=\log_4 7-\log_4 2=\log_4\frac72$$