The inverse of the function y = frac{10^x - 1 | vedclass

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

Question

(D) Given the function $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} + 1$. Subtract $1$ from both sides: $y - 1 = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$.

Vedclass · Accepted Answer

$\frac{1}{2} \log_{10} \left(\frac{y}{2-y}\right)$

Vedclass · Answer

(D) Given the function $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} + 1$. Subtract $1$ from both sides: $y - 1 = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$. Let $u = 10^x$. Then $10^{-x} = \frac{1}{u}$. So,$y - 1 = \frac{u - 1/u}{u + 1/u} = \frac{u^2 - 1}{u^2 + 1}$. Let $Y = y - 1$. Then $Y(u^2 + 1) = u^2 - 1$. $Yu^2 + Y = u^2 - 1 \implies Y + 1 = u^2(1 - Y)$. $u^2 = \frac{1 + Y}{1 - Y} = \frac{1 + (y - 1)}{1 - (y - 1)} = \frac{y}{2 - y}$. Since $u = 10^x$,we have $10^{2x} = \frac{y}{2 - y}$. Taking $\log_{10}$ on both sides: $2x = \log_{10} \left(\frac{y}{2 - y}\right)$. Therefore,$x = \frac{1}{2} \log_{10} \left(\frac{y}{2 - y}\right)$.

The inverse of the function $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} + 1$ is $x =$

Similar Questions

If $g$ is the inverse of $f$ and $f'(x) = \frac{1}{1 + x^5}$,then $g'(x) =$

If $g(x)$ is the inverse function of $f(x)$ and $f^{\prime}(x) = \frac{1}{1+x^4}$,then $g^{\prime}(x)$ is

Let $f: X \rightarrow Y$ be an invertible function. Show that the inverse of $f^{-1}$ is $f$,i.e.,$\left(f^{-1}\right)^{-1}=f$.

State with reason whether the following function has an inverse: $f: \{1,2,3,4\} \rightarrow \{10\}$ with $f = \{(1,10), (2,10), (3,10), (4,10)\}$.

If $f(x) = \frac{2x - 1}{x + 5}$ $(x \ne -5)$,then $f^{-1}(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module