ln X + ln X^2 = 3

Help I'm lost again =/

Also

Log X + Log(x - 21) = 2

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- Jul 24th 2007, 12:27 PMJonathanEyoonLogarithmic 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 PMCaptainBlack
- Jul 24th 2007, 12:43 PMJonathanEyoon
- Jul 24th 2007, 12:50 PMCaptainBlack
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 PMKrizalid
- Jul 24th 2007, 01:33 PMJonathanEyoon
- Jul 24th 2007, 01:34 PMJonboy
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 PMJonathanEyoon
- Jul 24th 2007, 01:43 PMJonboy
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 PMJonathanEyoon
- Jul 25th 2007, 06:17 AMtopsquark
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