$\log_[x]{81} - \log_[2x]{36}=2$

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- Feb 7th 2007, 11:19 PMperashfind x
$\log_[x]{81} - \log_[2x]{36}=2$

- Feb 7th 2007, 11:22 PMperash
how i can use latex

- Feb 8th 2007, 12:15 AMearboth

Hello,

have a look here: http://www.mathhelpforum.com/math-he...-tutorial.html

EB - Feb 8th 2007, 01:26 AMearboth
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 :D 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! - Feb 8th 2007, 09:38 AMperashfind x,y if
[tex]\sqrt{x}+y=3[\math]

[\sqrt{y}+x=5] - Feb 8th 2007, 01:26 PMtopsquark
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 - Feb 11th 2007, 01:26 PMoceanA=1/2bh solve for b