If cosh ^{-1} x = 2 log _e(sqrt{2}+1),then x= | vedclass

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

Question

(D) We have,$\cosh ^{-1} x = 2 \log _e(\sqrt{2}+1)$.Using the logarithmic property $n \log a = \log a^n$,we get:$\cosh ^{-1} x = \log _e(\sqrt{2}+1)^2

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) We have,$\cosh ^{-1} x = 2 \log _e(\sqrt{2}+1)$.Using the logarithmic property $n \log a = \log a^n$,we get:$\cosh ^{-1} x = \log _e(\sqrt{2}+1)^2$.We know that $\cosh ^{-1} x = \log _e(x + \sqrt{x^2-1})$.Therefore,$\log _e(x + \sqrt{x^2-1}) = \log _e(2 + 1 + 2\sqrt{2}) = \log _e(3 + 2\sqrt{2})$.Comparing both sides,we have $x + \sqrt{x^2-1} = 3 + 2\sqrt{2}$.Let $x = 3$,then $\sqrt{x^2-1} = \sqrt{9-1} = \sqrt{8} = 2\sqrt{2}$.Thus,$3 + 2\sqrt{2} = 3 + 2\sqrt{2}$,which satisfies the equation.Hence,$x = 3$.

If $\cosh ^{-1} x = 2 \log _e(\sqrt{2}+1)$,then $x=$

Similar Questions

Evaluate: $\frac{\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}} - 1}{\sqrt{5 + 2\sqrt{6}}}$

The value of $\sqrt{12\sqrt{5} + 2\sqrt{55}}$ is:

The real part of the principal value of $2^{-i}$ is

If $\sqrt{x + iy} = \pm(a + ib)$,then $\sqrt{-x - iy}$ is equal to

The imaginary part of $i^{i}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module