can somebody help me with this?
$$2=2^{1-x}+(\frac{2}{3})^{1-x}$$
I guess I somehow have to isolate this 1-x term and then use the ln. But I don't get how..
Thanks in advance!
Mostly irrelevant side quest: I have the (paid) Wolframalpha app which is supposed to show step-by-step solutions. However, 9/10 times it says "step-by-step solution unavailable". Anyone knows what is up with that?
We have that
$$f(x)=2^{1-x}+\left(\frac{2}{3}\right)^{1-x} \implies f'(x)=-2^{1-x}\log 2+\left(\frac{2}{3}\right)^{1-x}\log \frac32 \\\implies f''(x)=2^{1-x}\log^2 2+\left(\frac{2}{3}\right)^{1-x}\log^2 \frac32>0 $$
therefore $f(x)$ is convex and $f(x)=2$ has at most $2$ solutions which are by inspection $x=1$ and $x=2$.