The number of solutions of the equation ln(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 $\l

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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 $\ln x = t$. For $\ln x$ to be defined,$x > 0$. For $\ln(\ln x)$ to be defined,$\ln x > 0$,so $x > 1$. Also,for $\log_x e$ to be defined,$x > 0$ and $x 
eq 1$.Substituting $t = \ln x$,we get $\ln t = \frac{1}{t}$,or $t \ln t = 1$.Let $f(t) = t \ln t$. We want to find the number of solutions for $f(t) = 1$ where $t > 0$.$f'(t) = \ln t + 1$. Setting $f'(t) = 0$ gives $t = 1/e$.At $t = 1/e$,$f(1/e) = (1/e) \ln(1/e) = -1/e \approx -0.368$.As $t 	o 0^+$,$f(t) 	o 0$. As $t 	o \infty$,$f(t) 	o \infty$.Since $f(t)$ is continuous and strictly increasing for $t > 1/e$,and $f(1) = 0  1$,by the Intermediate Value Theorem,there exists exactly one solution for $t$ in the interval $(1, e)$.Since $t = \ln x$ is a one-to-one function,there is exactly one solution for $x$.

The number of solutions of the equation $\ln(\ln x) = \log_x e$ is:

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