Question

If $\log_3 2, \log_3(2^x-5), \log_3(2^x-\frac{7}{2})$ are in an arithmetic progression, then the value of x is equal to _________.

Hint:

When solving logarithmic equations, always remember to check your final solutions against the domain of the original logarithms. Solutions that make any argument less than or equal to zero are extraneous and must be discarded.

Answer 1

Correct Answer: 3

If three terms \(A,B,C\) are in arithmetic progression, then \[ 2B=A+C \] Applying this condition: \[ 2\log_3(2^x-5)=\log_3 2+\log_3\!\left(2^x-\frac{7}{2}\right) \] Using logarithmic properties: \[ 2\log_3 a=\log_3 a^2,\qquad \log_3 p+\log_3 q=\log_3(pq) \] \[ \log_3\bigl((2^x-5)^2\bigr)=\log_3\!\left(2\left(2^x-\frac{7}{2}\right)\right) \] Since \(\log_3\) is a one-to-one function, equate the arguments: \[ (2^x-5)^2=2(2^x)-7 \] Let \(y=2^x\). Then, \[ (y-5)^2=2y-7 \] \[ y^2-10y+25=2y-7 \] \[ y^2-12y+32=0 \] Factoring: \[ (y-4)(y-8)=0 \] \[ \Rightarrow y=4 \quad \text{or} \quad y=8 \] Substituting back: \[ 2^x=4 \Rightarrow x=2 \] \[ 2^x=8 \Rightarrow x=3 \] Step 2: Check domain of logarithms All logarithmic arguments must be positive: \[ 2^x-5>0 \Rightarrow 2^x>5 \] \[ 2^x-\frac{7}{2}>0 \Rightarrow 2^x>3.5 \] Thus, the stricter condition is: \[ 2^x>5 \] Checking values: \[ x=2 \Rightarrow 2^2=4 \;(\text{invalid}) \] \[ x=3 \Rightarrow 2^3=8 \;(\text{valid}) \] Answer: \(\boxed{3}\)