The number of solutions of log_{4}(x - 1) = l | vedclass

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(B) Given equation: $\log_{4}(x - 1) = \log_{2}(x - 3)$Using the base change formula $\log_{a^n}(b) = \frac{1}{n} \log_{a}(b)$,we have $\log_{4}(x - 1

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$1$

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(B) Given equation: $\log_{4}(x - 1) = \log_{2}(x - 3)$Using the base change 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)$$\log_{2}(x - 1) = 2 \log_{2}(x - 3)$$\log_{2}(x - 1) = \log_{2}((x - 3)^2)$$x - 1 = (x - 3)^2$$x - 1 = x^2 - 6x + 9$$x^2 - 7x + 10 = 0$$(x - 5)(x - 2) = 0$Thus,$x = 5$ or $x = 2$.Check the domain: For $\log_{2}(x - 3)$ to be defined,$x - 3 > 0$,so $x > 3$.For $x = 2$,$x - 3 = -1$,which is not allowed.For $x = 5$,$x - 3 = 2 > 0$,which is valid.Therefore,there is only $1$ solution.

The number of solutions of $\log_{4}(x - 1) = \log_{2}(x - 3)$ is:

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