1. ## Logarithmic Equations Problem

ln X + ln X^2 = 3

Help I'm lost again =/

Also

Log X + Log(x - 21) = 2

2. 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

3. 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

4. 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

5. Originally Posted by JonathanEyoon
$\displaystyle \log x + \log(x - 21) = 2$
Do the same that CaptainBlack's previous post.

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

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

7. 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.

8. 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!

9. 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!

10. 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

11. 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