2^5=32 how do I find the power with only 2 and 32?

Your function will basically look like

2^x = 32

To solve this, take the natural logarithm of both sides, i.e.,

x(ln 2) = ln (32)

and solving for "x"

x = 5

You can also use common logarithm instead of natural logarithm, i.e.

x(log 2) = log (32)

and come up with the same answer.

I am just more comfortable in using natural logarithm.

Hope this helps.

you have to take logs

2^x=32
log2^x=log32
xlog2=log32
x=log32/log2

that should give you 5

the power = log( 32 )/log( 2) = 5

you should know the powers of 2 up to 2^10 = 1024, so you "look up" what you need in your memory.

with a calculator you can solve
2^x = 32
log 2^x = log 32
x log 2 = log 32
x = log 32 / log 2

n(a million) = 5 n(2) = n(a million) +2 = 7 n(3) = n(2) +3 =10 n(4) = n(3) +4 =14 the specific formulation would would desire to be those 2 statements taken jointly: n(ok) = 5 for ok=a million n(ok) = n(ok-a million) + ok for ok=2,3,4,...

26x = 32

x log 2 = log 32

x = log 32 / log 2

x = 5 ----------( using any log base )

You could use logarithms, or if you don't have a calculator, you could do it the simple way.

[32] /2 {Count this as the first line, or 1}
=16 /2 {the second line, and it's still not the number, so divide it too}
=8 /2 {the third}
=4 /2 {the fourth}
=[2] {the fifth, and it's the number you want, so for 2^x=32, x evaluates as 5}

x = the power of 2

2^x = 32
2^x = 2^5
x = 5