The number of solutions of the equation log_{ | vedclass

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(B) Given equation: $\log_{7}(2^{x} - 1) + \log_{7}(2^{x} - 7) = 1$For the logarithms to be defined,we must have $2^{x} - 1 > 0$ and $2^{x} - 7 > 0$,w

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$1$

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(B) Given equation: $\log_{7}(2^{x} - 1) + \log_{7}(2^{x} - 7) = 1$For the logarithms to be defined,we must have $2^{x} - 1 > 0$ and $2^{x} - 7 > 0$,which implies $2^{x} > 7$.Using the property $\log_{b}(m) + \log_{b}(n) = \log_{b}(mn)$,we get:$\log_{7}((2^{x} - 1)(2^{x} - 7)) = 1$Taking the antilogarithm:$(2^{x} - 1)(2^{x} - 7) = 7^{1} = 7$Let $t = 2^{x}$. Then:$(t - 1)(t - 7) = 7$$t^{2} - 8t + 7 = 7$$t^{2} - 8t = 0$$t(t - 8) = 0$So,$t = 0$ or $t = 8$.Since $t = 2^{x}$,we have $2^{x} = 0$ (which has no solution) or $2^{x} = 8$.$2^{x} = 2^{3} \implies x = 3$.Checking the condition $2^{x} > 7$: for $x = 3$,$2^{3} = 8 > 7$,which is valid.Thus,there is only $1$ solution.

The number of solutions of the equation $\log_{7}(2^{x} - 1) + \log_{7}(2^{x} - 7) = 1$ is:

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