Log's and i'm lost

[reehlgirl18]

The Directions say to do these without a calculator.
*Fin the Exact value for each....

log8(1\8)=? what's the trick i forgot.

 

[stapel]

Assuming you mean "log-base-eight of one-eighth" (that is, "log₈(1/8)" or "\(\displaystyle \log_8{\left(\frac{1}{8}\right)}\)"), a good first step would be to express "one-eighth" in terms of 8 raised to some power. :wink:

Eliz.

 

[Unco]

The Directions say to do these without a calculator.
*Fin the Exact value for each....

log8(1\8)=? what's the trick i forgot.

Let \(\displaystyle \mbox{y = \log_8{\left(\frac{1}{8}\right)}\)

Take the exponential base 8 of both sides

\(\displaystyle \mbox{ 8^y = \frac{1}{8}}\)

\(\displaystyle \mbox{ \Rightarrow y = ?}\)

 

[Mrspi]

The Directions say to do these without a calculator.
*Fin the Exact value for each....

log8(1\8)=? what's the trick i forgot.

I'd rather call it a "technique" rather than a trick....set the log equal to x:

log<SUB>8</SUB> (1/8) = x

Change to exponential form, using the fact that if
log<SUB>b</SUB> a = x, then b<SUP>x</SUP> = a

If log<SUB>8</SUB> (1/8) = x, then 8<SUP>x</SUP> = 1/8

We can write 1/8 as 8<SUP>-1</SUP>. So,
8<SUP>x</SUP> = 8<SUP>-1</SUP>

and the exponents must be equal. Thus,
x = -1