The number of solution(s) of the equation ln( | vedclass

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(B) Given equation: $\ln(\ln x) = \log_x e$Using the property $\log_x e = \frac{1}{\ln x}$,the equation becomes $\ln(\ln x) = \frac{1}{\ln x}$.Let $t

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$1$

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(B) Given equation: $\ln(\ln x) = \log_x e$Using the property $\log_x e = \frac{1}{\ln x}$,the equation becomes $\ln(\ln x) = \frac{1}{\ln x}$.Let $t = \ln x$. Then the equation is $\ln t = \frac{1}{t}$,or $t \ln t = 1$.Let $f(t) = t \ln t$. We want to find the number of solutions for $f(t) = 1$ where $t > 0$ and $\ln x$ is defined (i.e.,$x > 0$ and $\ln x > 0$,so $t > 0$).For $t \in (0, 1]$,$f(t) = t \ln t \le 0$,so there is no solution.For $t > 1$,$f(t)$ is a strictly increasing function from $0$ to $\infty$.Since $f(1) = 0$ and $\lim_{t 	o \infty} f(t) = \infty$,by the Intermediate Value Theorem,there exists exactly one value $t_0 > 1$ such that $f(t_0) = 1$.Since $t = \ln x$,$x = e^{t_0}$ is the unique solution.Thus,the number of solutions is $1$.

The number of solution$(s)$ of the equation $\ln(\ln x) = \log_x e$ is -

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