<s>[INST] If $(2^b-1+c^b)^a = (2^a-1+c^a)^b,$ then is it true that $a=b?$ 
Question: Assume that $a,b,c$ are real numbers such that
$$1\leq a\leq b, \quad\text{and}\quad  0<c<1.$$
  If we have 
  $$(2^b-1+c^b)^a = (2^a-1+c^a)^b,$$
  then is it true that 
  $$a=b?$$

I can only prove that the statement is true if $a$ is an integer. 
In particular, since 
$$(2^b-1+c^b)^a = (2^a-1+c^a)^b,$$
by manipulation, we have 
$$2^a = (2^b-1+c^b)^{\frac{a}{b}} -c^a+1. $$
Since LHS is an even natural number, so is RHS.
This forces $a=b.$
However, I have no idea how to prove it for real number $a.$
 [/INST] Let
$$
f(a,b,c)=a\cdot\ln\Big(2^b-1+c^b\Big)-b\cdot\ln\Big(2^a-1+c^a\Big)
$$
Obviously your statement is equal to finding roots of $f$.
Now, its easy to see that $f(a,b,1)=0$, so we take the derivative:
$$
\frac{\partial f}{\partial c}= \frac{ab}{c(2^b-1+c^b)(2^a-1+c^a)}(2^ac^b-2^bc^a+c^a-c^b)
$$
Notice that the last term can be estimated as 
$$
2^ac^b-2^bc^a+c^a-c^b\leq2^bc^b-2^bc^a+c^a-c^b=(2^b-1)(c^b-c^a)<0
$$
due to $1\leq a<b$ and $0\leq c \leq 1$. Thus, $f$ is strictly decreasing in $c$ and hence has no other roots other than $c=1$, for $a\neq b$.
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