# Solve log 3 log10 y = -2

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

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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.

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Homework Helper
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?

Staff Emeritus