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

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 $\ln x = t$. 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:

Similar Questions

The quadratic equation with rational coefficients,whose one root is $\sqrt{5}$,is:

Solve the given two equations and select the correct answer from the given options.
$I.$ $12x + 3y = 14$
$II.$ $4x + 2y = 16$

If $\alpha, \beta$ are the roots of $x^2 - 2x + 4 = 0$,then $\alpha^5 + \beta^5$ is equal to

One root of the equation $3x^2 - 10x + 3 = 0$ is $\frac{1}{3}$. Find the other root.

When $a = -5, b = -6$ and $c = 10$,find the value of $\frac{a^{3} + b^{3} + c^{3} - 3abc}{ab + bc + ca - a^{2} - b^{2} - c^{2}}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module