The value of k for which the equation ln ( k ln x )= ln x has a unique solution, is

Question

The value of k for which the equation ln ( k ln x )= ln x has a unique solution, is (A) e (B) (1/ e ) (C) (1/ e 2) (D) 2 e

Tardigrade Admin · Accepted Answer

ln ( k ln x )= ln x ⇒ ln x =( x / k ) ....(1) The line y =( x / k ) becomes tangent to the curve y = ln x at x = e for k = e. So, the solution of the equation (1) is x = k text at k = c

Q. The value of $k$ for which the equation $ln (k ln x) = ln x$ has a unique solution, is

$e$

$\frac{1}{e}$

$\frac{1}{e ^{2}}$

$2 e$

$ln (k ln x) = ln x \Rightarrow ln x = \frac{x}{k}$ ....(1)
The line $y = \frac{x}{k}$ becomes tangent to the curve $y = ln x$ at $x = e$ for $k = e$ .
So, the solution of the equation (1) is
$x = k at k = c$

Q. The value of k for which the equation ln(klnx)=lnx has a unique solution, is

e

e1​

e21​

2e

ln(klnx)=lnx⇒lnx=kx​ ....(1) The line y=kx​ becomes tangent to the curve y=lnx at x=e for k=e. So, the solution of the equation (1) is x=k at k=c

Q. The value of $k$ for which the equation $ln (k ln x) = ln x$ has a unique solution, is

$e$

$\frac{1}{e}$

$\frac{1}{e ^{2}}$

$2 e$

$ln (k ln x) = ln x \Rightarrow ln x = \frac{x}{k}$ ....(1)
The line $y = \frac{x}{k}$ becomes tangent to the curve $y = ln x$ at $x = e$ for $k = e$ .
So, the solution of the equation (1) is
$x = k at k = c$