- #1

DeanBH

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- 0

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.

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- Thread starter DeanBH
- Start date

- #1

DeanBH

- 82

- 0

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

rocomath

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

Power Rule: [tex]\log y^3=-2 \leftrightarrow 3\log y=-2[/tex]

Logarithmic to Exponential form: [tex]\log_a B=m \leftrightarrow a^m=B[/tex]

Power Rule: [tex]\log y^3=-2 \leftrightarrow 3\log y=-2[/tex]

Logarithmic to Exponential form: [tex]\log_a B=m \leftrightarrow a^m=B[/tex]

Last edited:

- #3

DeanBH

- 82

- 0

so

3logy = -2

logy^3 = -2

log10^-2 = y^3

y=10^(-2/3)

3logy = -2

logy^3 = -2

log10^-2 = y^3

y=10^(-2/3)

- #4

rocomath

- 1,755

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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:

- #5

HallsofIvy

Science Advisor

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3 log y= -2 so log y= -2/3. Now, the example you showed says that y= 10

- #6

DeanBH

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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?

- #7

cristo

Staff Emeritus

Science Advisor

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

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