$\log_[x]{81} - \log_[2x]{36}=2$
Hello,
have a look here: http://www.mathhelpforum.com/math-he...-tutorial.html
EB
Hello,
I assume that you know $\displaystyle x=e^{\ln(x)}$
and
$\displaystyle \log_{b}{(x)}=\frac{\ln(x)}{\ln(b)}$
$\displaystyle \log_{x}{(81)}-\log_{2x}{(36)}=2 \Longleftrightarrow \frac{\ln(81)}{\ln(x)}-\frac{\ln(36)}{\ln(2x)}=2$
With $\displaystyle \ln(2x)=\ln(x)+\ln(2)$ the LCD is: $\displaystyle \ln(x) \cdot (\ln(x)+\ln(2))$
$\displaystyle \frac{4\ln(3) \cdot (\ln(x)+\ln(2))}{\ln(x) \cdot (\ln(x)+\ln(2))}-\frac{(2\ln(2)+2\ln(3)) \cdot \ln(x)}{\ln(x) \cdot (\ln(x)+\ln(2))}=2$
After a few simple transformations you should get:
(From this point the Latex doesn't work properly. I'll post the rest as plain text)
$\displaystyle (\ln(x))^2+\ln(4/3)*\ln(x) - 2*\ln(3)*\ln(2)=0$
I'm using now the logarithms in decimal form. Substitute: y=ln(x) and you'll get:
$\displaystyle y^2+0.287682y-5.5079002=0 \Longleftrightarrow y=1.0986123 \vee y=-1.3862943$
Now re-substitute using $\displaystyle x=e^y$ and you'll get: $\displaystyle x = 3 \ \vee \ x = \frac{1}{4}$.
Probably there is a more elegant and easier way to solve this equation - but I couldn't find it.
EB
PS.: Obviously the Latex is back to duty!
If you have a new question, please post it in a new thread.
$\displaystyle \sqrt{x} + y = 3 $ (Your "\" on the last LaTeX bracket is backward.)
$\displaystyle \sqrt{y} + x = 5$
Take the first equation:
$\displaystyle y = 3 - \sqrt{x}$
Then insert this into the second equation:
(*) $\displaystyle \sqrt{3 - \sqrt{x}} + x = 5$
$\displaystyle \sqrt{3 - \sqrt{x}} = 5 - x$ <-- Square both sides.
$\displaystyle 3 - \sqrt{x} = 25 - 10x + x^2$
$\displaystyle \sqrt{x} = -x^2 + 10x - 22$ <-- Squaring again.
$\displaystyle x = x^4 - 20x^3 + 144x - 440x + 484$
$\displaystyle x^4 - 20x^3 + 144x^2 - 441x + 484 = 0$
I can't find a way to factor this thing, so I'm going to suggest a numerical solution. I get that x = 4 and x = 5.8 or so. (The other two solutions don't work.) Thus y = 1 and y = 0.591681084 or so, respectively.
-Dan