If$ a+b-1=1+\frac{ln(2^a-1)}{ln4}+\frac{ln(2^b-1)}{ln4}$ then $a=b$?

Question

If $$a+b-1=1+\frac{ln(2^a-1)}{ln4}+\frac{ln(2^b-1)}{ln4}$$ where $a,b>0$ are real numbers and ln is $log_e$, then is a=b?

Answer 1

Let $f(x) = x - \log_4 \left(2^{x+2}-4\right)$. Then the given relation is $f(a)+f(b) = 0$.

Now $f(x)$ has a minimum at $x=1$, which is $f(1) = 0$. So $f(x) > 0$ for all other values of $x>0$. Hence for the relation to be satisfied, $a=b=1$.

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Updated: to show the minimum without calculus, note that

$f(x) \ge 0 \iff x \ge \log_4(2^{x+2}-4) \iff 4^{x} \ge 2^{2+x}-4 \iff y^2\ge 4y - 4 \iff (y-2)^2 \ge 0$ where $y = 2^x$

and this also $\implies f(x) = 0 \iff 2^x=2 \iff x = 1$