What is the number of solutions for the equat | vedclass

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(B) Given the equation: $\log_4(x - 1) = \log_2(x - 3)$.First,note the domain constraints: $x - 1 > 0 \implies x > 1$ and $x - 3 > 0 \implies x > 3$.

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$1$

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(B) Given the equation: $\log_4(x - 1) = \log_2(x - 3)$.First,note the domain constraints: $x - 1 > 0 \implies x > 1$ and $x - 3 > 0 \implies x > 3$. Thus,$x > 3$.Using the change of base formula,$\log_4(x - 1) = \frac{\log_2(x - 1)}{\log_2(4)} = \frac{\log_2(x - 1)}{2}$.Substituting this into the equation: $\frac{\log_2(x - 1)}{2} = \log_2(x - 3)$.$\log_2(x - 1) = 2 \log_2(x - 3) = \log_2((x - 3)^2)$.Equating the arguments: $x - 1 = (x - 3)^2$.$x - 1 = x^2 - 6x + 9$.$x^2 - 7x + 10 = 0$.$(x - 2)(x - 5) = 0$.So,$x = 2$ or $x = 5$.Checking the domain $x > 3$: $x = 2$ is rejected,and $x = 5$ is accepted.Therefore,there is only $1$ solution.

What is the number of solutions for the equation $\log_4(x - 1) = \log_2(x - 3)$?

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