# Solve log 3 log10 y = -2

1. May 11, 2008

### DeanBH

how do logs work when it's like this.

3 log10 y = -2

log10 y = a
10^a = y

so it's 10^-2 = y

but where does the 3 go, and why does it go there. i'm not sure.

2. May 11, 2008

### rocomath

FFR (for future reference) :p ln is base e and log by itself, is implied base 10.

Power Rule: $$\log y^3=-2 \leftrightarrow 3\log y=-2$$

Logarithmic to Exponential form: $$\log_a B=m \leftrightarrow a^m=B$$

Last edited: May 11, 2008
3. May 11, 2008

### DeanBH

so

3logy = -2
logy^3 = -2
log10^-2 = y^3
y=10^(-2/3)

4. May 11, 2008

### rocomath

Your third line is wrong, but your final answer is correct. I'm not sure what you were doing, but you wouldn't take the log of the right side then drop the log on the left.

Last edited: May 11, 2008
5. May 11, 2008

### HallsofIvy

Staff Emeritus
Another way to do that problem is to just divide both sides by 3 at the start:
3 log y= -2 so log y= -2/3. Now, the example you showed says that y= 10-2/3 as before.

6. May 12, 2008

### DeanBH

should be 10^-2 = y^3 without log?

7. May 12, 2008

### cristo

Staff Emeritus
Whilst your first point is always true, your second point is not. I know people (myself included) who sometimes write log(x) to be base e. One should always check the conventions of the book that one is using.