Question

How do you solve `4^(2x-5)=32`?

Answer 1

Real solution `x = 15/4`

Complex solutions `x = 15/4+(kpi)/(2 ln 2) i` for any integer `k`

Explanation:

Note that `4 = 2^2` and `32 = 2^5`.

So we have:

`2^(4x-10) = (2^2)^(2x-5) = 4^(2x-5) = 32 = 2^5`

The function `2^x` is one to one as a real valued function from `RR` to `(0, oo)`.

Hence the only real solution is given by:

`4x-10 = 5`

Hence:

`x = 15/4`

If we are interested in other complex solutions, note that:

`2^((2kpii)/ln 2) = e^(2kpii) = (e^(2pii))^k = 1^k = 1" "` for any integer `k`

Hence the given equation has solutions given by:

`4x-10 = 5+(2kpii)/ln 2`

Hence:

`x = 15/4+(kpi)/(2 ln 2) i`

Answer 2

`15/4`

Explanation:

In order to solve, you can manipulate the bases of the exponents to find a value for x.

In this question, you can manipulate the base of 4 to be `2^2` and 32 to be `2^5`. It should look like this:
`2^(2(2x-5))=2^5`

Now that the bases equivalent, you can take a logarithm with a base of 2 to both sides of the equation to cancel out the exponents.
`log_2(2^(2(2x-5)))=log_2(2^5)`

`log_2(2)=1` so you can cancel out the log functions on both sides. This gives you:
`2(2x-5)=5`

Now, just solve like a normal algebraic equation.
`4x-10=5`
`4x=15`
`x=15/4`