# Math Help - Logarithms

1. ## Logarithms

Hi,
I'm not sure if I got these problems right. Can someone help me?

1. write "ln5" as a multiple of a common logarithm.
2. write "ln(x^2-1)/x^3)" as a sum, difference and/or constant multiples of logarithms. I got "ln(x^2-1)-3lnx" for this.
3. Simplify "log(9/300)". I got "log3"
4. Simplify -8+e^(ln(x^3)). I have no idea how to solve this one

I'll really appreciate it if someone can answer these questions. Thanks!

Isabel

2. Originally Posted by cuteisa89
Hi,
I'm not sure if I got these problems right. Can someone help me?

1. write "ln5" as a multiple of a common logarithm.
ln 5 = log 5/ln e
2. write "ln(x^2-1)/x^3)" as a sum, difference and/or constant multiples of logarithms. I got "ln(x^2-1)-3lnx" for this.
ln(x^2-1) - ln (x^3) = ln [(x+1)(x-1)]- 3ln x=ln(x+1)+ln(x-1)-3*ln x
3. Simplify "log(9/300)". I got "log3"
9/300=300
Thus,
log 9/300 = log 300 = log 3*10^2 = log 3 + log 10^2=log 3+2
4. Simplify -8+e^(ln(x^3)). I have no idea how to solve this one
Note
e^(ln x)=x for all positive x.
Now,
ln x^3 = 3ln x
Thus,
e^(ln(x^3))=e^(3ln x)=(e^(ln x))^3 (by the exponent laws).
But then,
(e^ln(x))^3=x^3.

Or, an easier way.
Let y=x^3.
Thus,
e^(ln y) = y= x^3.

This is Mine 5-th Post!!!

3. Originally Posted by cuteisa89
1. write "ln5" as a multiple of a common logarithm.
i'm not exactly sure what they expect here.

2. write "ln(x^2-1)/x^3)" as a sum, difference and/or constant multiples of logarithms. I got "ln(x^2-1)-3lnx" for this.
you are right so far, but you can go further:

ln(x^2-1)-3lnx = ln(x + 1)(x - 1) - 3lnx = ln(x + 1) + ln(x - 1) - 3lnx

3. Simplify "log(9/300)". I got "log3"
is this log to the base 3 or the normal log to the base 10? (either way, log3 is wrong)

log(9/300) = log(3/100) = log3 - log100 = log3 - 2 (if it's log to the base 10)

4. Simplify -8+e^(ln(x^3)). I have no idea how to solve this one
-8+e^(ln(x^3)) = -8 + x^3 = x^3 - 8

note that e^ln(x) = x

4. Originally Posted by ThePerfectHacker
ln 5 = log 5/ln e

ln(x^2-1) - ln (x^3) = ln [(x+1)(x-1)]- 3ln x=ln(x+1)+ln(x-1)-3*ln x

9/300=300
Thus,
log 9/300 = log 300 = log 3*10^2 = log 3 + log 10^2=log 3+2

Note
e^(ln x)=x for all positive x.
Now,
ln x^3 = 3ln x
Thus,
e^(ln(x^3))=e^(3ln x)=(e^(ln x))^3 (by the exponent laws).
But then,
(e^ln(x))^3=x^3.

Or, an easier way.
Let y=x^3.
Thus,
e^(ln y) = y= x^3.

This is Mine 5-th Post!!!
now i know how Soroban feels when he replies to a question only to see I beat him to it