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

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- February 8th 2007, 12:19 AMperashfind x
$\log_[x]{81} - \log_[2x]{36}=2$

- February 8th 2007, 12:22 AMperash
how i can use latex

- February 8th 2007, 01:15 AMearboth

Hello,

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

EB - February 8th 2007, 02:26 AMearboth
Hello,

I assume that you know

and

With the LCD is:

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)

I'm using now the logarithms in decimal form. Substitute: y=ln(x) and you'll get:

Now re-substitute using and you'll get: .

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! - February 8th 2007, 10:38 AMperashfind x,y if
[tex]\sqrt{x}+y=3[\math]

[\sqrt{y}+x=5] - February 8th 2007, 02:26 PMtopsquark
If you have a new question, please post it in a new thread.

(Your "\" on the last LaTeX bracket is backward.)

Take the first equation:

Then insert this into the second equation:

(*)

<-- Square both sides.

<-- Squaring again.

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 - February 11th 2007, 02:26 PMoceanA=1/2bh solve for b