Prove or refute that $\log_5(0.7-2^x)=\log_2(0.7-5^x)$ if and only if $x=-1$ I have been trying to prove that the following two functions intersect at a single point $(-1,-1)$:


$f(x)=\log_5(0.7-2^x)$
$g(x)=\log_2(0.7-5^x)$


It is sufficient to prove that:


If $x&gt;-1$, then $g(x)&gt;f(x)$
If $x&lt;-1$, then $g(x)&lt;f(x)$


I believe it should be relatively easy to prove that they don't have more than two intersection points (though I fail to see how this might help in proving that they only have a single intersection point).

Can anyone help proving or refuting that?

Here is the graph of both functions:
It is sufficient to prove that $f'&lt;g'$:

\begin{align}f'(x)&lt;g'(x)&amp;\iff-\frac{2^x}{.7-2^x}\frac{\log2}{\log5}&lt;-\frac{5^x}{.7-5^x}\frac{\log5}{\log2}\\
&amp;\iff\frac{2^x}{.7-2^x}\frac{.7-5^x}{5^x}&gt;\frac{\log^25}{\log^22}=:k\\
&amp;\iff\frac{.7\cdot5^{-x}-1}{.7\cdot2^{-x}-1}&gt;k\\
&amp;\iff(.7\cdot5^{-x}-1)&gt;(.7\cdot2^{-x}-1)k\\
&amp;\iff2^{-x}k-5^{-x}&lt;\frac{10}7(k-1)\\
\end{align}

Thus we are showing that $h(x):=2^{-x}k-5^{-x}$ has a particular upper bound. Since it is a difference of exponentials, we already know it has one root and is upper bounded. The derivative $h'(x)=2^{-x}k\log 2-5^{-x}\log 5$, which is zero when $x=x^\circ:=-\log_{5/2}\log_25$. Then $$h(x)\le 2^{-x^\circ}k-5^{-x^\circ}\approx5.805&lt;6.273\approx\frac{10}7(k-1).$$

(Sorry I had to resort to numerical evaluations, but this is a rigorous proof, since with a good enough evaluation of the two constants you can bound one above the other.)
Let $f(x) = \log_5(0.7-2^x)-\log_2(0.7-5^x)$; $f(x)$ is continuous and differentiable for $x&lt;\log_2(0.7)\approx-0.51$, so you can apply Rolle's Theorem. By absurd, suppose there exist two numbers $x_1$ and $x_2$, such that $F(x_1)=f(x_2)=0$, then by the Rolle's Theorem, there exists a number $c$ in $]x_1;x_2[$, such that $f'(c)=\frac{2^c\ln(2)}{(2^c-0.7)\ln(5)}-\frac{5^c\ln(5)}{(5^c-0.7)\ln(2)}=0 \Rightarrow \frac{2^c\ln(2)}{(2^c-0.7)\ln(5)}=\frac{5^c\ln(5)}{(5^c-0.7)\ln(2)}$ but this equation has no real solution! Absurd conclusion $\to$ there is only 1 solution $(x=-1)$.