The number of solutions of log_4(x - 1) = log | vedclass

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(B) Given equation: $\log_4(x - 1) = \log_2(x - 3)$First,define the domain: $x - 1 > 0 \Rightarrow x > 1$ and $x - 3 > 0 \Rightarrow x > 3$. Thus,$x >

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$1$

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(B) Given equation: $\log_4(x - 1) = \log_2(x - 3)$First,define the domain: $x - 1 > 0 \Rightarrow x > 1$ and $x - 3 > 0 \Rightarrow x > 3$. Thus,$x > 3$.Using the change of base formula $\log_{a^n}(b) = \frac{1}{n}\log_a(b)$,we have $\log_4(x - 1) = \frac{1}{2}\log_2(x - 1)$.So,$\frac{1}{2}\log_2(x - 1) = \log_2(x - 3)$.Multiplying by $2$,we get $\log_2(x - 1) = 2\log_2(x - 3) = \log_2((x - 3)^2)$.Equating the arguments: $x - 1 = (x - 3)^2$.$x - 1 = x^2 - 6x + 9$.$x^2 - 7x + 10 = 0$.$(x - 5)(x - 2) = 0$.This gives $x = 5$ or $x = 2$.Since the domain requires $x > 3$,$x = 2$ is an extraneous solution.Therefore,the only valid solution is $x = 5$.The number of solutions is $1$.

The number of solutions of $\log_4(x - 1) = \log_2(x - 3)$ is:

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