# Solve `2^(16^x)=16^(2^x)` for `x` .   ``

`2^((16)^(x))` = `16^((2)^(x))`

Using logarithm:

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

`=>` `16^x (log(2)) = 2^x (log (16))`

`=>16^x = 2^x * ((log (16))/(log (2)))`

`=> 16^x = 4 * 2^x`

`=>log (16^x) = log (4* 2^x)`

`=>x (log (16)) = log (2^x) + (log (4))`

`=> x(log (16)) - x(log (2)) = log (4)`

We then take x as common and separate x from the logarithms. So, we can divide log (4) by (log (16) - log(2)).

`=> x = log (4)/ (log (16)- log (2))`

`=> x = 2/3`

This uses the following logarithm rules:

1. `log (a^b) = b(log (a))`
2. `log (a*b) = log (a) + log (b)`

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sciencesolve | Certified Educator

You need to solve for x the equation:

`2^(16^x) = 16^(2^x)`

You need to write the base 16 as a multiple of two,` 2^4 = 16` . The reason of this action is to create the same base to the right side, such that:

`2^(16^x) = (2^4)^(2^x)`

`2^(16^x) = 2^(4*2^x)`

`16^x = 4*2^x => (2^4)^x = 2^2*2^x => 2^(4x) = 2^(2+x)`

Equating the exponents, yields:

`4x = 2 + x => 4x - x = 2 => 3x = 2 => x = 2/3`

Hence, evaluating the solution to the given equation, yields ` x = 2/3.`