# find x

• Feb 7th 2007, 11:19 PM
perash
find x
$\log_[x]{81} - \log_[2x]{36}=2$
• Feb 7th 2007, 11:22 PM
perash
how i can use latex
• Feb 8th 2007, 12:15 AM
earboth
Quote:

Originally Posted by perash
how i can use latex

Hello,

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

EB
• Feb 8th 2007, 01:26 AM
earboth
Quote:

Originally Posted by perash
$\log_[x]{81} - \log_[2x]{36}=2$

Hello,

I assume that you know $x=e^{\ln(x)}$
and
$\log_{b}{(x)}=\frac{\ln(x)}{\ln(b)}$

$\log_{x}{(81)}-\log_{2x}{(36)}=2 \Longleftrightarrow \frac{\ln(81)}{\ln(x)}-\frac{\ln(36)}{\ln(2x)}=2$

With $\ln(2x)=\ln(x)+\ln(2)$ the LCD is: $\ln(x) \cdot (\ln(x)+\ln(2))$

$\frac{4\ln(3) \cdot (\ln(x)+\ln(2))}{\ln(x) \cdot (\ln(x)+\ln(2))}-\frac{(2\ln(2)+2\ln(3)) \cdot \ln(x)}{\ln(x) \cdot (\ln(x)+\ln(2))}=2$

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)

$(\ln(x))^2+\ln(4/3)*\ln(x) - 2*\ln(3)*\ln(2)=0$
I'm using now the logarithms in decimal form. Substitute: y=ln(x) and you'll get:

$y^2+0.287682y-5.5079002=0 \Longleftrightarrow y=1.0986123 \vee y=-1.3862943$

Now re-substitute using $x=e^y$ and you'll get: $x = 3 \ \vee \ x = \frac{1}{4}$.

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 AM
perash
find x,y if
[tex]\sqrt{x}+y=3[\math]
[\sqrt{y}+x=5]
• Feb 8th 2007, 01:26 PM
topsquark
Quote:

Originally Posted by perash
[tex]\sqrt{x}+y=3[\math]
[\sqrt{y}+x=5]

If you have a new question, please post it in a new thread.

$\sqrt{x} + y = 3$ (Your "\" on the last LaTeX bracket is backward.)
$\sqrt{y} + x = 5$

Take the first equation:
$y = 3 - \sqrt{x}$

Then insert this into the second equation:
(*) $\sqrt{3 - \sqrt{x}} + x = 5$

$\sqrt{3 - \sqrt{x}} = 5 - x$ <-- Square both sides.

$3 - \sqrt{x} = 25 - 10x + x^2$

$\sqrt{x} = -x^2 + 10x - 22$ <-- Squaring again.

$x = x^4 - 20x^3 + 144x - 440x + 484$

$x^4 - 20x^3 + 144x^2 - 441x + 484 = 0$

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 PM
ocean
A=1/2bh solve for b
$\log_[x]{81} - \log_[2x]{36}=2$
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