Find $x$ if $2^{16^x} = 16^{2^x}$.
If $x = 0$ we have $2=16$.
$8$ times or $(2^3)$ difference and if $x = 1$ we have $65536=256$.
$256$ or $2^8$ difference. I can't see anything useful in figuring this out.
I have a feeling $x$ must be a fraction. I know only know about logarithm. I haven't learned anything about the number $e$ or $ln$ yet.
$\endgroup$ 63 Answers
$\begingroup$$16^{2^x}=(2^4)^{2^x}=2^{4\cdot 2^x}=2^{2^2\cdot 2^x}=2^{2^{x+2}}$. So$$2^{16^x}=16^{2^x}\iff 2^{16^x}=2^{2^{x+2}}\iff$$ $$\iff 16^x=2^{x+2}\iff 2^{4x}=2^{x+2} \iff 4x=x+2 \iff x=\frac{2}{3}. $$
$\endgroup$ $\begingroup$$$2^{16^x} = 2^{(2^4)^x} = 2^{2^{4x}}$$
$$16^{2^x} = (2^4)^{2^x} = 2^{4\times 2^x} = 2^{2^{x+2}}$$
Now equate the two expressions.
$\endgroup$ $\begingroup$If you take the $\log_2$ of both sides you end up with\begin{align*} \log_2(2^{16^x}) &= 16^x \log_2(2) = 16^x = 2^x\cdot8^x\\ \log_2(16^{2^x}) &= 2^x\log_2{16} = 2^x\cdot4 \end{align*}From this you get that $2^x\cdot 4 = 2^x \cdot 8^x \implies 8^x = 4 \implies x =\log_{8}(4) = 2/3.$
($\log_8(4) = \log_8(2^2) = 2\log_8(2) = 2\cdot1/3$)
$\endgroup$
