Question

How do you solve `20^(9-4n)=5`?

Answer 1

`n=(9-log_20 5)/4`

Explanation:

We know that :

`x^y=z iff y=log_xz`

so :

`20^(9-4n)=5 iff 9-4n=log_20 5 iff 4n=9-log_20 5 iff`

`n=(9-log_20 5)/4`

Answer 2

`n =9/4-1/4(log5/log20)`

`n = 2.116`

Explanation:

In a question like this you first have to decide whether you will solve it as an exponential equation, or use logs.

`5` is certainly not a power of `20`, so logs are indicated.

`20^(9-4n) =5" "larr` find log of both sides

`log20^(color(blue)(9-4n)) =log5" "larr` apply the power law

`color(blue)((9-4n))log 20 = log5" "larr` isolate the factor with `n`

`9-4n = log5/log20" "larr` re-arrange to get `4n` positive

`9-log5/log20 = 4n" "larr` divide by 4

`1/4(9-log5/log20) = n`

`n = 9/4 - 1/4(log5/log20)`

`n = 2.116` (to 3 d.p.)