The number of solutions of \(\log_4(x-1)=\log_2(x-3)\). (a) 3 (b) 1 (c) 2 (d) 0

Answer: option (b) \(\mathbf{1}\)

Solution:

Given: \(\log_4(x-1)=\log_2(x-3)\)

\(\log_{2^2}(x-1)=\log_2(x-3)\)

\(\Rightarrow\frac{1}{2}\log_{2}(x-1)=\log_2(x-3)\)       \(\mathbf{ [∵ \log_{a^m}(x)=\frac{1}{m} \log_a(x)]}\)

\(\Rightarrow \log_{2}(x-1)=2\log_2(x-3)\)

\(\Rightarrow \log_{2}(x-1)=\log_2(x-3)^2\)           \(\mathbf{ [∵ n\log_ax=\log_ax^n]}\)

\(\Rightarrow (x-1)=(x-3)^2\)            \(\mathbf{ [∵\log_a(x)=\log_a(y)\implies x=y]}\)

\(\Rightarrow (x-1)=x^2+9-6x\)         \(\mathbf{ [∵(a-b)^2=a^2+b^2-2ab]}\)

\(\Rightarrow x^2-7x+10=0\)

\(\Rightarrow x^2-2x-5x+10=0\)

\(\Rightarrow x(x-2)-5(x-2)=0\)

\(\Rightarrow (x-2)(x-5)=0\)

\(\Rightarrow   (x-2)=0, (x-5)=0\)

\(\Rightarrow x=2, x=5\)

The equality in \(\log_4(x-1)=\log_2(x-3)\) holds only if \(x-3>0 \implies x>3\)

So only value of \(x\) which satisfy the equation is \(x=5\).

Therefore, the number of solution is \(1\).

Hence, option (b) \(\mathbf{1}\) is correct answer.