If _32, _3(2^x-5), _3(2^x-72) are in an arithmetic progression, then the value of x is equ…

Question

If $\log_32, \log_3(2^x-5), \log_3\left(2^x-\dfrac72\right)$ are in an arithmetic progression, then the value of $x$ is equal to \_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

$2\log_3(2^x-5) = \log_32+\log_3\left(2^x-	frac72ight) \Rightarrow (2^x-5)^2 = 2\left(2^x-	frac72ight).$

Let $t=2^x$: $t^2-12t+32=0 \Rightarrow (t-4)(t-8)=0 \Rightarrow t=4$ or $8$, so $x=2$ or $3$.

If $x=2$: $2^x-5=-1<0$, invalid for the logarithm. So $x=3$.

Solutions · 1