# Logarithmic Equations Problem

• Jul 24th 2007, 12:27 PM
JonathanEyoon
Logarithmic Equations Problem
ln X + ln X^2 = 3

Help I'm lost again =/

Also

Log X + Log(x - 21) = 2
• Jul 24th 2007, 12:34 PM
CaptainBlack
Quote:

Originally Posted by JonathanEyoon
ln X + ln X^2 = 3

Help I'm lost again =/

I'll assume you mean:

$\displaystyle \ln(x) + \ln(x^2) =3$,

Well:

$\displaystyle \ln(x) + \ln(x^2) = \ln(x \times x^2)=\ln(x^3)=3\ln(x)$,

so $\displaystyle \ln(x)=1$, and so $\displaystyle x=e$.

RonL
• Jul 24th 2007, 12:43 PM
JonathanEyoon
Quote:

Originally Posted by CaptainBlack
I'll assume you mean:

$\displaystyle \ln(x) + \ln(x^2) =3$,

Well:

$\displaystyle \ln(x) + \ln(x^2) = \ln(x \times x^2)=\ln(x^3)=3\ln(x)$,

so $\displaystyle \ln(x)=1$, and so $\displaystyle x=e$.

RonL

Uhhmmm... could you explain it a little easier
• Jul 24th 2007, 12:50 PM
CaptainBlack
Quote:

Originally Posted by JonathanEyoon
Uhhmmm... could you explain it a little easier

The logarithm property that was the original reason the were invented:

$\displaystyle \ln(A)+\ln(B)=\ln(A\times B)$

so:

$\displaystyle \ln(x) + \ln(x^2)=\ln(x \times x^2)=\ln(x^3)$

The second logarithm property needed is:

$\displaystyle \ln(x^n) = n\ln(x)$

(which is realy just the previous property in disguise),

so:

$\displaystyle \ln(x) + \ln(x^2)=\ln(x \times x^2)=\ln(x^3)=3\ln(x)$

Therefore we have:

$\displaystyle 3 \ln(x)=3$

and so:

$\displaystyle \ln(x)=1$.

But the number who's log (to any base) is 1 is the base itself, and since we have logs to the base $\displaystyle e$ here; $\displaystyle x=e$

RonL
• Jul 24th 2007, 12:53 PM
Krizalid
Quote:

Originally Posted by JonathanEyoon
$\displaystyle \log x + \log(x - 21) = 2$

Do the same that CaptainBlack's previous post.
• Jul 24th 2007, 01:33 PM
JonathanEyoon
Quote:

Originally Posted by Krizalid
Do the same that CaptainBlack's previous post.

The equation should come out to x^2 - 21x -2 right?
• Jul 24th 2007, 01:34 PM
Jonboy
You are almost correct Jonathan (hey that's my name too!).

So we have: $\displaystyle logx\,+\,log(x\,-\,21)\,=\,2$

Which goes to: $\displaystyle log(x(x\,-\,21))\,=\,2$

Now raise both to a base of 10: $\displaystyle x^2\,-\,21x\,=\,100$

You can solve the quadratic, make sure to check that both solutions work.
• Jul 24th 2007, 01:41 PM
JonathanEyoon
Quote:

Originally Posted by Jonboy
You are almost correct Jonathan (hey that's my name too!).

So we have: $\displaystyle logx\,+\,log(x\,-\,21)\,=\,2$

Which goes to: $\displaystyle log(x(x\,-\,21))\,=\,2$

Now raise both to a base of 10: $\displaystyle x^2\,-\,21x\,=\,100$

You can solve the quadratic, make sure to check that both solutions work.

Thanks so much I totally forgot that Log has a natural base of 10. Appreciate it!
• Jul 24th 2007, 01:43 PM
Jonboy
No prob. Make sure that when you do problem like these you do them step by step and don't rush like me... it'll cause a slight but critical error. Luckily I caught myself. Good luck!
• Jul 24th 2007, 01:47 PM
JonathanEyoon
Quote:

Originally Posted by Jonboy
No prob. Make sure that when you do problem like these you do them step by step and don't rush like me... it'll cause a slight but critical error. Luckily I caught myself. Good luck!

I need help with one more problem =/

ln(4x - 2) - ln4 = -ln(x - 2)

I'm having trouble setting up so i can solve =/

Nevermind I solved it
• Jul 25th 2007, 06:17 AM
topsquark
Quote:

Originally Posted by JonathanEyoon
I need help with one more problem =/

ln(4x - 2) - ln4 = -ln(x - 2)

I'm having trouble setting up so i can solve =/

Nevermind I solved it

In case anyone else needs the help on it:
$\displaystyle ln(4x - 2) - ln4 = -ln(x - 2)$

$\displaystyle ln \left ( \frac{4x - 2}{4} \right ) = ln[(x - 2)^{-1}]$

$\displaystyle ln \left ( \frac{4x - 2}{4} \right ) = ln \left ( \frac{1}{x - 2} \right )$ <-- Take both sides to the e power

$\displaystyle \frac{4x - 2}{4} = \frac{1}{x - 2}$

$\displaystyle (4x - 2)(x - 2) = 4$

$\displaystyle 4x^2 - 10x + 4 = 4$

$\displaystyle 4x^2 - 10x = 0$

$\displaystyle 2x(2x - 5) = 0$

So either
$\displaystyle 2x = 0$ ==> $\displaystyle x = 0$
or
$\displaystyle 2x - 5 = 0$ ==> $\displaystyle x = \frac{5}{2}$

Substitution into the original equation show that both solutions work.

-Dan