If y = cosh(log x) + sinh(log x), prove that y^n = 0 for n > 1.
Understand the Problem
The question is asking us to prove that y raised to the power n equals 0 for n greater than 1, given the equation y = cosh(log x) + sinh(log x). The approach involves manipulating the mathematical expression to understand its behavior as n exceeds 1.
Answer
$y^n \neq 0$ for all real $x > 0$ and $n > 1$.
Answer for screen readers
$y^n \neq 0$ for all real $x > 0$ and $n > 1$.
Steps to Solve
- Rewrite the hyperbolic functions using exponents
Recall the definitions of hyperbolic cosine and hyperbolic sine: $$ \cosh(x) = \frac{e^x + e^{-x}}{2} $$ $$ \sinh(x) = \frac{e^x - e^{-x}}{2} $$ Substituting $x = \log(x)$, we have: $$ \cosh(\log(x)) = \frac{e^{\log(x)} + e^{-\log(x)}}{2} $$ $$ \sinh(\log(x)) = \frac{e^{\log(x)} - e^{-\log(x)}}{2} $$
- Simplify the expressions for y
Using the properties of logarithms, $e^{\log(x)} = x$ and $e^{-\log(x)} = \frac{1}{x}$: $$ y = \frac{x + \frac{1}{x}}{2} + \frac{x - \frac{1}{x}}{2} $$ This simplifies further: $$ y = \frac{2x}{2} = x $$
- Prove $y^n = 0$ for $n > 1$
If we need to prove $y^n = 0$ for $n > 1$, we substitute $y$ back into the expression: $$ y^n = x^n $$ This will equal 0 if $x = 0$. However, for all real $x$ greater than 0, $x^n \neq 0$.
- Conclusion on the existence of solutions
Since $y = x$, we conclude that $y^n = 0$ is only true if $x = 0$, which is outside the domain for $\log(x)$. Therefore, $y^n \neq 0$ for $n > 1$ under the given constraints.
$y^n \neq 0$ for all real $x > 0$ and $n > 1$.
More Information
This result illustrates that while certain mathematical relationships might seem to suggest that $y^n$ can equal zero, the logarithm function restricts $x$ to positive values, preventing $y^n$ from ever being zero for $n > 1$.
Tips
- Assuming $y^n = 0$ could hold for $n > 1$ without considering the domain of $x$.
- Forgetting that $y = x$ must be evaluated within the context of real numbers where $x$ cannot be zero.
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