This is the question which I am referring to
if log(base 3 of 2), log (base 3 of (2^(x)-5)), log(base 3 of (2^(x)-(7/2)) are in AP, find the value of x
My try and answer
I used the equation 2b= a+c , where a, b, c are in AP and solved for x so I got the answer log (base 2 of (6+(2)^(1/2)) and log (base 2 of (6-(2)^(1/2)) which comes out to be 2.89 and 2.17 respectively Is my way to solve question correct? Correct answer is 3.
Read, understand and justify the following (assuming we have $\;2^x-\frac72\;$ in the last term):
$$\log_32\;,\;\;\log_3(2^x-5)=\log_32+d\;,\;\;\log_3\left(2^x-\frac72\right)=\log_32+2d$$
and also (we drop the subindex $\;3\;$ )
$$2\log(2^x-5)=\log2+\log\left(2^x-\frac72\right)=\log(2^{x+1}-7)\implies$$
$$2^{2x}-10\cdot2^x+25=2\cdot2^x-7$$
Substitute accordingly and get the quadratic $\;t^2-12t+32=(t-8)(t-4)\;$ , and end the exercise.