If log₃2, log₃(2ˣ − 5), log₃(2ˣ − 7/2) are in arithmetic progression, then the value of x is equal to —
Explanation
\text{Given } \log_3 2,\ \log_3(2^x-5),\ \log_3\!\left(2^x-\frac{7}{2}\right)\ \text{are in AP.}
\text{Let } a=\log_3 2,\quad b=\log_3(2^x-5),\quad c=\log_3\!\left(2^x-\frac{7}{2}\right).
\text{Since } a,b,c \text{ are in AP, } 2b=a+c.
\begin{align*}
2\log_3(2^x-5) &= \log_3 2 + \log_3\!\left(2^x-\frac{7}{2}\right)\\
\log_3(2^x-5)^2 &= \log_3\!\left(2\left(2^x-\frac{7}{2}\right)\right)\\
(2^x-5)^2 &= 2^{x+1}-7.
\end{align*}
\text{Put } y=2^x:
</span><br><spanstyle="font−size:14pt;font−family:arial,helvetica,sans−serif;">(y−5)2=2y−7⇒y2−12y+32=0.</span><br><spanstyle="font−size:14pt;font−family:arial,helvetica,sans−serif;">
</span><br><spanstyle="font−size:14pt;font−family:arial,helvetica,sans−serif;">y=4 or y=8.</span><br><spanstyle="font−size:14pt;font−family:arial,helvetica,sans−serif;">
So 2x=4⇒x=2 or 2x=8⇒x=3.
\text{Check domain: }2^x-5>0. For }x=2,\ 2^2-5=-1\ (\text{invalid}). Hence discard x=2.
\boxed{x=3}
