Logarithmic Equations Problem

**JonathanEyoon** · July 24th 2007, 12:27 PM

ln X + ln X^2 = 3

Help I'm lost again =/

Also

Log X + Log(x - 21) = 2

**CaptainBlack** · July 24th 2007, 12:34 PM

Originally Posted by JonathanEyoon

ln X + ln X^2 = 3

Help I'm lost again =/

I'll assume you mean:

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

Well:

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

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

RonL

**JonathanEyoon** · July 24th 2007, 12:43 PM

Originally Posted by CaptainBlack

I'll assume you mean:

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

Well:

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

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

RonL

Uhhmmm... could you explain it a little easier

**CaptainBlack** · July 24th 2007, 12:50 PM

Originally Posted by JonathanEyoon

Uhhmmm... could you explain it a little easier

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

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

so:

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

The second logarithm property needed is:

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

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

so:

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

Therefore we have:

$3 \ln(x)=3$

and so:

$\ln(x)=1$ .

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

RonL

**Krizalid** · July 24th 2007, 12:53 PM

Originally Posted by JonathanEyoon

$\log x + \log(x - 21) = 2$

Do the same that CaptainBlack's previous post.

**JonathanEyoon** · July 24th 2007, 01:33 PM

Originally Posted by Krizalid

Do the same that CaptainBlack's previous post.

The equation should come out to x^2 - 21x -2 right?

**Jonboy** · July 24th 2007, 01:34 PM

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

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

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

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

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

**JonathanEyoon** · July 24th 2007, 01:41 PM

Originally Posted by Jonboy

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

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

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

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

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!

**Jonboy** · July 24th 2007, 01:43 PM

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!

**JonathanEyoon** · July 24th 2007, 01:47 PM

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

**topsquark** · July 25th 2007, 06:17 AM

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

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

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

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

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

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

$4x^2 - 10x = 0$

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

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

Substitution into the original equation show that both solutions work.

-Dan

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