1. ## Log Help

Seems to be a lot of log help going on tonight.. here's one that im not sure where to start..

${{6}}^{{{6}{x}-{3}}}={{5}}^{{{4}{x}-{6}}}$

also,

${\log}_{{{4}}}{\left({{x}}^{{2}}\right)}-{\log}_{{{4}}}{\left({x}+{5}\right)}={4}$
Now on this on i figured out that i need to convert to exponential form this gives me:

$\displaystyle 4^4=x^2/(x+5)$

then i multiply out:

$\displaystyle 256x+1280=x^2$

rearrange:
$\displaystyle x^2-256x-1280=0$

now i figure punch it into the quadratic formula and call it a day? the program i'm putting it into for HW says it's only 1/2 right if i put in $\displaystyle (256+sqrt(70656))/2$ and yes i tried $\displaystyle (256-sqrt(70656))/2$ since it is +/-... ideas?

oh and by the way, how do i put square root symbols in these messages?

2. Originally Posted by xsavethesporksx
${{6}}^{{{6}{x}-{3}}}={{5}}^{{{4}{x}-{6}}}$
Hint:

Let k = log(5) / log(6)

(6x - 3) / (4x - 6) = k

3. [QUOTE=xsavethesporksx;475365]Seems to be a lot of log help going on tonight.. here's one that im not sure where to start..

${{6}}^{{{6}{x}-{3}}}={{5}}^{{{4}{x}-{6}}}$

Take the logarithm (either "common" or "natural"- whichever is simpler on your calculator) of both sides to get (6x-3)log(6)= (4x- 6)log(5) so that 6log(6)x- 3log(6)= 4log(5)x- 6log(5). Solve that linear equation for x. That is the same as saying $\displaystyle \frac{6x- 3}{4x- 6}= \frac{log(5)}{log(6)}$ as Wilmer gives.

also,

${\log}_{{{4}}}{\left({{x}}^{{2}}\right)}-{\log}_{{{4}}}{\left({x}+{5}\right)}={4}$
Now on this on i figured out that i need to convert to exponential form this gives me:

$\displaystyle 4^4=x^2/(x+5)$

then i multiply out:

$\displaystyle 256x+1280=x^2$

rearrange:
$\displaystyle x^2-256x-1280=0$

now i figure punch it into the quadratic formula and call it a day? the program i'm putting it into for HW says it's only 1/2 right if i put in $\displaystyle (256+sqrt(70656))/2$ and yes i tried $\displaystyle (256-sqrt(70656))/2$ since it is +/-... ideas?
The quadratic equation gives two roots- and since squaring or adding 5 to the negative root gives a positive value in the square root, bothare valid solutions.

oh and by the way, how do i put square root symbols in these messages?
The best thing to do is to learn "LaTex". For example, [ math ]\int_{-\infty}^\infty e^{-x^2}dx= \sqrt{2\pi} [ /math ] without the spaces inside [ ] gives $\displaystyle \int_{-\infty}^\infty e^{-x^2}dx= \sqrt{2\pi}$.

Clicking on any such formula will show you the code to produce it. There is a tutorial here:
http://www.mathhelpforum.com/math-help/latex-help/