Logarithm inverses... confusion!

I am having trouble understanding this. Probably because the distance course I'm doing kind of jumps steps and doesn't explain.

I am just kind of stuck trying to inverse log functions.

It gives me:

y = -log(base 5)(-x)

how do I invert that?

I know it'd be something like:

x = 5^-y

.... but is it: -x = 5^-y? then how would I solve for y? I'm just confused and since it's distance I don't really have a teacher readily available to ask. I thank you in advance for your help. Really frustrates me when I don't understand something, and then not being able to ask someone really makes it worse!

Thanks

You already had it solved for y (when you have y alone on one side, and some function of x on the other side, it's solved for y)

y = -log₅(-x)

Now if you wish to take minus sign away from front of log, just remember this identity:
a log(b) = log(b^a)

y = -1 * log₅(-x)
y = log₅(-x^(-1))
y = log₅(-1/x)

====================

Solving for x, we'd get:

y = -log₅(-x)
-y = log₅(-x)
5^(-y) = 5^log₅(-x)
5^(-y) = -x
x = -5^(-y)

How is y = -(1/5)^x not correct? --graph the two equations and they are images of one another reflected through the line y = x? --why'd that get a thumbs down, man?

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You've answered it yourself !

-x=5^-y i.e. x=-5^-y

You aren't solving for y, you are solving for x. In other words you want to know what x is given what y=-log(5)(-x) is. So you are finding the inverse of -log(5)(-x).

It looks like you're right to me--

x = -5^-y --> y = -5^-x or y = -(1/5)^x

Just like the inverse of y = ln(base e)(x) --> y = e^x.