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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(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 -

Similar Questions

Let $\log _a b=4$ and $\log _c d=2$,where $a, b, c, d$ are natural numbers. Given that $b-d=7$,the value of $c-a$ is:

If $\log_{5} a \cdot \log_{a} x = 2$,then $x$ is equal to

If ${a^x} = b$,${b^y} = c$,and ${c^z} = a$,then the value of $xyz$ is:

If $n = (2020)!$,then the value of $\frac{1}{\log _{2} n} + \frac{1}{\log _{3} n} + \frac{1}{\log _{4} n} + \ldots + \frac{1}{\log _{2020} n}$ is equal to:

If $\log _2 x + \log _4 x + \log _8 x + \log _{16} x = \frac{25}{36}$ and $x = 2^k$,then $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module