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1, log_9 sqrt(3^x+48), log_9(3^x-8/3) in AP: Find x | JEE

JEE Maths question with a full step-by-step solution.

Question
If 1, log⁡93x+48, log⁡9(3x−83)1,\ \log_9\sqrt{3^x+48},\ \log_9\left(3^x-\frac83\right) are in A.P., then the value of xx is
A99
B66
C22correct
D44
Solution
rguments in G.P.:
9, 3x+48, 3x−83 are in G.P.9,\ \sqrt{3^x+48},\ 3^x-\frac83\ \text{are in G.P.}
 Step 2: The G.P. condition gives
(3x+48)2=9(3x−83) ⇒ 3x+48=9⋅3x−24.\left(\sqrt{3^x+48}\right)^2=9\left(3^x-\frac83\right)\ \Rightarrow\ 3^x+48=9\cdot 3^x-24.
 Step 3:
8⋅3x=72 ⇒ 3x=9=32 ⇒ x=2.8\cdot 3^x=72\ \Rightarrow\ 3^x=9=3^2\ \Rightarrow\ x=2.
 Correct answer: (3)
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