what are the zeroes of $f(x) = 2^x + 3^x - 5^x?$

Question

1) How do I find the number of zeroes of such exponential equations?

2) Is there any easier way to manually plot such sum of exponential functions?

Answer 1

Notice that $f(x)=0 \iff f(x)5^{-x}=0$. Therefore, it suffices to study $g(x)=f(x)5^{-x}$

$$g(x)= \left(\frac{2}{5}\right)^x+\left(\frac{3}{5}\right)^x-1 $$ The function $g$ is decreasing as a sum of two decreasing functions. Hence, $g$ has at most one root. However, $\lim_{x\rightarrow -\infty} g(x)=\infty$ and $\lim_{x\rightarrow \infty}g(x)=-1$, therefore $g$ has exactly one root.

Also by trial and error $f(1)=0$, so $r=1$ is the unique root for $f$.

Answer 2

At $x=0$, we know that the formula becomes $3^0+2^0-5^0=1+1-1=1$, and when $x<0$ $3^x>5^x$ and $2^x>5^x$. Therefore, $2^x+3^x>5^x$, and there are no roots for $x<0$. For $x>0$, we know that $2^x<3^x<5^x$, so at some point $5^x=2^x+3^x$ and at point $x$ there will be a zero. That just happens to be $x=1$ since $2^x+3^x-5^x=2^1+3^1-5^1=2+3-5=5-5=0$.