Logarithms!

Using logarithms to solve this problem, we really only need the knowledge of 2 logarithmic properties to solve it.

Property A : $\log_a b^c=c \times (\log_a b)$

Property B : $\log_c ab=\log_c a + \log_c b$

Now, to solve it:

2^16^x=16^2^x

Apply log to both sides:

log 2^16^x=log 16^2^x

Using Property A :

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

$16^{x} \times (log 2)=2^x \times (log 2^4)$

$16^{x} \times (log 2)=(2^x \times 4)(log 2)$

Dividing both sides by log 2:

$16^{x}=(2^x \times 4)$

Using Property A on the left and Property B on the right:

$x \times (log 16)=(log 2^x + log 4)$

Using Property A even more:

$x \times (log 2^4)=(log 2^x + log 2^2)$

$(x \times 4) \times (log 2)=x \times (log 2) + 2 \times (log 2)$

Dividing each side by log 2:

$4x=x + 2$

Solving the simple equation:

$3x=2$

$\boxed{\boxed{\boxed{\boxed{x=\frac {2}{3}}}}}$

Hope you get the problem!