How do logs work with multiple bases and negative exponents?

[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.

 

[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: $$\log_a B=m \leftrightarrow a^m=B$$

 

[DeanBH]

so

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

 

[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.

 

[HallsofIvy]

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.

 

[DeanBH]

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.

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

 

[cristo]

ln is base e and log by itself, is implied base 10.

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.