Question

If x = log (y +√y2 + 1 ) then y =

• A. tanh x
• B. coth x
• C. sinh x
• D. cosh x

Answer 1

The Correct Option is C

We are given the equation $x = \log(y + \sqrt{y^2 + 1})$ and need to express $y$ in terms of $x$.

1. Initial Transformation:
Starting with $x = \log(y + \sqrt{y^2 + 1})$, we exponentiate both sides:
$e^x = y + \sqrt{y^2 + 1}$

2. Isolating the Square Root:
$\sqrt{y^2 + 1} = e^x - y$

3. Squaring Both Sides:
$y^2 + 1 = (e^x - y)^2 = e^{2x} - 2ye^x + y^2$

4. Simplifying the Equation:
$y^2 + 1 = e^{2x} - 2ye^x + y^2$
$1 = e^{2x} - 2ye^x$

5. Solving for y:
$2ye^x = e^{2x} - 1$
$y = \frac{e^{2x} - 1}{2e^x} = \frac{e^x - e^{-x}}{2}$

6. Recognizing the Hyperbolic Function:
Recall that $\sinh x = \frac{e^x - e^{-x}}{2}$, so we conclude:
$y = \sinh x$

7. Verification:
To verify, let $y = \sinh x = \frac{e^x - e^{-x}}{2}$:
$\sqrt{y^2 + 1} = \cosh x = \frac{e^x + e^{-x}}{2}$
$y + \sqrt{y^2 + 1} = \sinh x + \cosh x = e^x$
$\log(y + \sqrt{y^2 + 1}) = \log(e^x) = x$

Final Answer:
The solution is ${\sinh x}$.