Logarithm equations

[Laura54321]

I really don't know how to do these equations:

1. log10(1 3/7)
log105=0.6990 and log107=0.8451 is what they give me.


2. log8(n-3)+log8(n+4)=1

 

[pka]

I cannot read your #1.

For #2
\(\displaystyle \L \begin{array}{l}
\log _8 (n - 3) + \log _8 (n + 4) = 1 \\
\log _8 \left[ {(n - 3)(n + 4)} \right] = 1 \\
(n - 3)(n + 4) = 8 \\
\end{array}.\)

You can now finish.

 

[Laura54321]

Thanks for your help on number 2! Do you just replace the 1 with the 8 or do you have to multiply it into the 1?


I'll re-write that other problem:

Use log105 = .6990 and log107 = .8451 to evaluate the expression.

log10(1 3/7)

I don't know how to figure out what I'm supposed to do to get numbers into 1 3/7. :?

 

[skeeter]

change the mixed number 1 and 3/7 to the improper fraction 10/7 ...

log(10/7) = log(10) - log(7) = 1 - .8451 = .1549

 

[stapel]

Do you just replace the 1 with the 8 or do you have to multiply it into the 1?

Has your class not covered the definition of "logarithms"? (If it had, it would have covered the way to convert between log equations and the equivalent exponential equations, is why I ask.)

Thank you!

Eliz.

 

[Laura54321]

My teacher didn't go into detail of how to convert logs into expontential form so I'm confused on how to do it.

 

[pka]

The base must be positive and if \(\displaystyle a > 0\) then the definition of logarithm is:
\(\displaystyle \L \log _b (a) = c\quad \mbox{if and only if}\quad b^c = a\)

Examples:
\(\displaystyle \begin{array}{rcl}
\log _4 (256) = x\quad & \Rightarrow & \quad 4^x = 256 \\
\log _b (81) = 4\quad & \Rightarrow & \quad b^4 = 81 \\
\log _{\left( {1/2} \right)} (x) = 8\quad & \Rightarrow & \quad x = \left( {1/2} \right)^8 \\
\end{array}\)

 

[Laura54321]

Thanks you so much! That helps a lot!