ln X + ln X^2 = 3
Help I'm lost again =/
Also
Log X + Log(x - 21) = 2
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
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.
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