Is it possible to convert $y = a^x + b^x$ to the form of $y = a \cdot b^x$?

Question

I don't think this is possible, but I wanted to ask people who know more than me.

Is it possible to convert $y = a^x + b^x$ to the form of $y = a \cdot b^x$?

Answer 1

I will interpret the question to mean:

For function $f_{a,b}(x) = a^x + b^x$, do parameters $\alpha$ and $\beta$ exist such that $f_{a,b}(x) =\alpha \beta^x$?

The answer is no.

Assume that such parameters do exist. Let $x = 0$, to get $\alpha = 2$ and then let $x = 1$ to get $\beta = \frac{a+b}2$. Now, take any other $x$ to derive a contradiction.

Edit: Of course, I assumed $a\neq b$. If $a = b$ then $a^x + b^x = 2a^x$. Conversely, take $\alpha = 2$ and $\beta = \frac{a+b}2$ as above and let $x = 2$. It immediately follows that $(a-b)^2 = 0 \implies a = b$. So, actually, more precise answer would be: "such $\alpha$ and $\beta$ exist if and only if $a = b$ and in that case, $\alpha = 2$, $\beta = a = b$."

Answer 2

$y=(a/b)^x*b^x+b^x$

$y= (a/b)^x*b^x+b^x = b^x((a/b)^x+1)$

This is really the closest to the form you want to get (reducing the $(a/b)^x+1$ to a constant would be impossible). Its a nice simplification depending on the ratio of a/b.

Example if $a>>b$ then $y = a^x$ if $a=b$ then $y=2a^x=2*b^x$ if $a<<b$ then $y=b^x$ which are the extreme values. If $b>a$ then y tends to $b^x$ but is slightly higher then. If $b<a$ y tends to a^x but a bit higher.