How do I find $\gamma$ such that $100^{\gamma} = \frac{1}{2} (144^{\gamma} + 64 ^{\gamma})$?
I know the solution is $\gamma=\frac{1}{2}$.
I have tried:
Unpacking the equation into products of prime numbers and eyeballing a common base.
Juggling the terms around with no particular outcome in mind to see if I recognized another form of the expression.
Applying log to both sides.
Substituting $3^2+4^2=5^2$ for the $5^2$ factor in $100^{\gamma}$.
My idea: Simplify first (I will use $x$)
\begin{align} 100^x &= \frac{1}{2}*(144^x+64^x)\\ 2 &= \frac{144^x + 64^x}{100^x}\\ 2 &= \left( \frac{144}{100} \right) ^x + \left( \frac{64}{100} \right) ^x\\ 2 &= \left( \frac{36}{25} \right) ^x + \left( \frac{16}{25} \right) ^x\\ \end{align}
Here we look for the easiest ways to solve. Noticing that both terms on the right (like @semiclassical said) are perfect squares would result in rational terms. This gives us the obvious $x = 0, \frac{1}{2}$
Finding the derivative of $f(x) = (\frac{36}{25})^x + (\frac{16}{25})^x $,
$$f'(x) = \left( \frac{36}{25} \right) ^x*\ln \left( \frac{36}{25} \right) + \left( \frac{16}{25} \right) ^x\ln \left( \frac{16}{25} \right)$$
Here, we notice that there is only one negative term in this, $\ln(16/25)$. Therefore, we can extrapolate that there will only be one zero.
Looking at the limit of $f(x)$, we see that it approaches $\infty$ on both sides, and we can pretty much intuitively see that it is strictly increasing when $x \gt$ critical point and strictly decreasing when $x \lt$ critical point.
Therefore, assuming that the critical point/minimum is below $y=2$, there can be a maximum of two solutions. We have found both of them, so we are done.