Question

How do you solve `32^n = 1` ?

How about `32^n = 2` and `32^n = 8` ?

Answer 1

`n=0`

Explanation:

The only real-valued solution is `n=0`

If `a` is any non-zero number then `a^0 = 1`.

[ Side note: `0^0` is probably best considered indeterminate ]

So if `a=32` then one solution is `n=0`.

There are no other real solutions, since `32^n` is strictly monotonically increasing, so one to one as a real valued function of real numbers.

For the example:

`32^n = 2`

Note that `32 = 2^5`. So we have:

`2^1 = 2 = 32^n = (2^5)^n = 2^(5n)`

Hence real solution `n = 1/5`

For the example:

`32^n = 8`

We have:

`2^3 = 8 = 32^n = 2^(5n)`

Hence real solution `n = 3/5`

Complex solutions

There are possible complex solutions too, given by:

`n = (2kpi)/ln 32 i" "` for any integer `k`

To see why, note that Euler's identity tells us:

`e^(ipi)+1 = 0`

That is:

`e^(ipi) = -1`

Hence:

`e^(2pii) = (-1)^2 = 1`

and for any integer `k`:

`e^(2kpii) = 1`

Also note that:

`32^n = (e^(ln 32))^n = e^(n ln 32)`

Hence if `n = (2kpii)/ln 32` for any integer `k`, then:

`32^n = 32^((2kpi)/ln 32 i) = e^(ln 32 * (2kpi)/ln 32 i) = e^(2kpii) = 1`

Similarly:

`32^(1/5+(2kpi)/ln 32 i) = 2`

`32^(3/5+(2kpi)/ln 32 i) = 8`

Answer 2

`0`

Explanation:

from the laws of powers

`x^0=1,AAx in RR`

so

`32^n=1=>n=0`