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

(A) Given the function $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$.Multiply the numerator and denominator by $10^x$:$y = \frac{10^{2x} - 1}{10^{2x} +

Vedclass · Accepted Answer

$\frac{1}{2} \log _{10}\left(\frac{1+x}{1-x}\right)$

Vedclass · Answer

(A) Given the function $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$.Multiply the numerator and denominator by $10^x$:$y = \frac{10^{2x} - 1}{10^{2x} + 1}$Now,solve for $x$ in terms of $y$:$y(10^{2x} + 1) = 10^{2x} - 1$$y \cdot 10^{2x} + y = 10^{2x} - 1$$1 + y = 10^{2x} - y \cdot 10^{2x}$$1 + y = 10^{2x}(1 - y)$$10^{2x} = \frac{1 + y}{1 - y}$Taking $\log_{10}$ on both sides:$2x = \log_{10} \left( \frac{1 + y}{1 - y} \right)$$x = \frac{1}{2} \log_{10} \left( \frac{1 + y}{1 - y} \right)$Replacing $y$ with $x$ to find the inverse function $f^{-1}(x)$:$f^{-1}(x) = \frac{1}{2} \log_{10} \left( \frac{1 + x}{1 - x} \right)$

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

Similar Questions

Let $f: X \rightarrow Y$ be an invertible function. Show that $f$ has a unique inverse.
(Hint: Suppose $g_{1}$ and $g_{2}$ are two inverses of $f$. Then for all $y \in Y$,$f \circ g_{1}(y) = I_{Y}(y) = f \circ g_{2}(y)$. Use the one-one property of $f$.)

If $y = f(x) = \frac{ax + b}{cx - a}$,then $x$ is equal to

Let $f(x) = \sin x + \cos x$ and $g(x) = x^2 - 1$. Then $g(f(x))$ is invertible for $x \in $

Let $g$ be the inverse function of $f$ and $f'(x) = \frac{x^{10}}{1 + x^2}$. If $g(2) = a$,then $g'(2)$ is equal to:

If $f(x) = (2x - 3\pi)^5 + \frac{4}{3}x + \cos x$ and $g$ is the inverse of $f$,then $g'(2\pi) = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module