Crazy Exponential

$\Rightarrow \large 2^{{16}^x} = 16^{{2}^x}$ $\Rightarrow \large 2^{{16}^x} = 2^{4\cdot{2}^x}$ $\text{Bases are same, we can equate powers.}$ $\Rightarrow \large 16^x = 2^2\cdot 2^x$ $\Rightarrow \large 2^{4x} = 2^{x+2}$ $\text{Again bases are same,we can equate powers.}$ $\Rightarrow \large 4x = x + 2$ $\large \boxed{x= 0.667}$

$\LARGE2^{16^{x}}=16^{2^{x}}$ $\LARGE2^{2^{4x}}=2^{4\times2^{x}}$ $\textbf{Will equate powers, as the bases are equal}$ $\LARGE2^{4x}=2^2\times2^{x}$ $\textbf{Divide by}\space\Large2^x$ $\LARGE2^{(4-1)\times x}=2^2$ $\textbf{Again bases are equal}$ $\Large3x=2$ $\Large\color{#D61F06}{\boxed{\boxed{\color{#3D99F6}{x=\frac23=0.66667}}}}$ $\LARGE\color{#D61F06}{or}$ $\LARGE2^{2^{4x}}=2^{4\times2^{x}}$ $\LARGE2^{2^{4x}}=2^{2^2\times2^{x}}$ $\LARGE2^{2^{4x}}=2^{2^{(x+2)}}$ $\textbf{Will equate powers, as the bases are equal}$ $\Large 4x=x+2$ $\Large3x=2$ $\Large\color{#D61F06}{\boxed{\boxed{\color{#3D99F6}{x=\frac23=0.66667}}}}$

So I took logarithms instead of using rules of indices, as I've been caught out before because apparently I don't know all the laws of indices.

So I'm working with log base 2 (not the standard base 10 or natural)

$log(2^{16^x}) = log(16^{2^x})$

by laws of logarithms is the same as:

$(16^x)log2 = (2^x)log16$

$16^x = 2^x * 4$

now take logarithms again (still in base 2):

$log(16^x)=log(2^x * 4)$

by rules of logarithms:

$xlog16=log(2^x) + log4$

$xlog16=xlog2+log4$

$4x=x+2$

now by basic algebra we get

$3x=2$

$x=2/3=0.666$

If 2^16^x = 16^2^x => 16^2/2^x = ln(16)/ln(2) = (16/2)^x = 8^x => xln(8)=ln(ln(16)/ln(2)) => x = ln(ln(16)/ln(2))/ln(8) => x = 2/3 or approx. 0.667

Just basic laws of indices

2^{16^x}=16^{2^x} --- (1)

Taking logarithms

16^x log 2=2^x log 16 ---- (2)

Therefore

8^x=(log 16/log 2)=4 ---- (3)

(2^3)^x=2^2 ------- (4)

i.e.,

2^{3x}=2^2 --- (5)

i.e.,

3x=2 => x=2/3=0.6666667