# If a + b &#x2212;<!-- − --> 1 = 1 + l n (

If $a+b-1=1+\frac{ln\left({2}^{a}-1\right)}{ln4}+\frac{ln\left({2}^{b}-1\right)}{ln4}$ then $a=b$?
where $a,b>0$ are real numbers and ln is $lo{g}_{e}$, then is a=b?
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furniranizq
Let $f\left(x\right)=x-{\mathrm{log}}_{4}\left({2}^{x+2}-4\right)$. Then the given relation is $f\left(a\right)+f\left(b\right)=0$
Now $f\left(x\right)$ has a minimum at $x=1$, which is $f\left(1\right)=0$. So $f\left(x\right)>0$ for all other values of $x>0$. Hence for the relation to be satisfied, $a=b=1$
Updated: to show the minimum without calculus, note that
$f\left(x\right)\ge 0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x\ge {\mathrm{log}}_{4}\left({2}^{x+2}-4\right)\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{4}^{x}\ge {2}^{2+x}-4\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{y}^{2}\ge 4y-4\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\left(y-2{\right)}^{2}\ge 0$
where $y={2}^{x}$
and this also $\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}f\left(x\right)=0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{2}^{x}=2\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x=1$