Question

$2 \times \sqrt[2]{180} {22y}^{2}$

Answer 1

It depends on what you mean by 'solve'. You can start substituting
values for x, and see which one works:

2^0 = 1
2^1 = 2
2^2 = 4
2^3 = 8 Success!

This doesn't work so well for something like 2^x = 9:

2^0 = 1
2^1 = 2
2^2 = 4
2^3 = 8 Too small
2^4 = 16 Too large



2^3.1 = 8.57
2^? = 9 It's somewhere between 3.1 and 3.5
2^3.5 = 11.31

Or, you can use logarithms.

By definition, if

x
b = a

then

log a = x
b

In this case, if

x
2 = 8

then

log 8 = x