Learning Logs: Solving y=.70+.59x

[Jemal]

I'm trying to teach myself logs for a statistics quiz... I know this is probably easy but i never really learned logs.

Homework Statement

I have to take the logarithm of both side of y=.70 + .59x

Homework Equations

n/a

The Attempt at a Solution

y=.70+.59x

So i get Logy= Log.70 + Log (.59x)
Logy = -.1549 + log.59x

I don't know what to do with the log(.59x) Can i separate the .59 and the x somehow?

 

[gb7nash]

y=.70+.59x

So i get Logy= Log.70 + Log (.59x)

No. You can't do this. We only know that log(ab) = log(a) + log(b). The correct equation is:

Log(y) = Log(.70+.59x)

What is it you're trying to do?

 

[Jemal]

No. You can't do this. We only know that log(ab) = log(a) + log(b). The correct equation is:

Log(y) = Log(.70+.59x)

What is it you're trying to do?

I just need to take the log of both sides and simplify.

So I get log(y) = log(.70+.59x)

I can't seem to find any laws of logs that let's me simplify log(.70+.59x)

Is that how the equation can be left?

 

[gb7nash]

Pretty much. There's not much else you can do with that equation.

 

[Jemal]

Pretty much. There's not much else you can do with that equation.

ok thanks a bunch

 

[Mark44]

I have to take the logarithm of both side of y=.70 + .59x

Why do you think you need to do this? What's the context from which this arises?