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

If $\alpha_1, \alpha_2$ and $\beta_1, \beta_2$ are the roots of the equations $ax^2 + bx + c = 0$ and $px^2 + qx + r = 0$ respectively,and the system of equations $\alpha_1 y + \alpha_2 z = 0$ and $\beta_1 y + \beta_2 z = 0$ has a non-zero solution,then:

If the equation $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$ has at least one solution,then $(b + a)$ can be equal to

Solve the given two equations and select the correct answer from the given options.
$I.$ $5x^2 + 3x - 14 = 0$
$II.$ $10y^2 - 3y - 27 = 0$

If the roots of the equation $ax^2 + x + b = 0$ are real,then the roots of the equation $x^2 - 4\sqrt{ab}x + 1 = 0$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module