You may also use the following method, avoiding logarithms, hence, you may write `1/16` as `1/2^4` .

Using the negative power property, `1/2^4 = 2^(-4).`

Substituting `2^(-4)` for `1/16` in the given equation, yields:

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

Using the property of bijective function, you may set the exponents equal, such that:

`x = -4`

**Hence, evaluating the solution to the given equation yields `x = -4.` **

`2^x = 1/16`

Take log on both sides.

`log(2^x) = log(1/16)`

We know the following related to log;

`log(a^n) = nloga`

`log(a/b) = loga-logb`

`log1 = 0`

`log(2^x) = log(1/16)`

`xlog2 = log1-log16`

`xlog2 = 0-log16`

`xlog2 = -log(2^4)`

`xlog2 = -4log2`

`x = -4`

*So the answer is x = -4*