# Solving a logarithmic equation

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• May 3rd 2010, 05:18 AM
alternative
Solving a logarithmic equation
Hi, could anyone help me with this, I'd be very greatful.Thanks..

20logx^1/2 + 7logx^3 - 3logx^2 = 0
4x 16x x/2
• May 3rd 2010, 05:38 AM
mathaddict
Quote:

Originally Posted by alternative
Hi, could anyone help me with this, I'd be very greatful.Thanks..

20logx^1/2 + 7logx^3 - 3logx^2 = 0
4x 16x x/2

hi

This is confusing . You might want to try the latex (a software to format equations) .

Try this ,

logarithms of x to base 3 can be typed as \log_3(x) ,
then wrap it with this $\displaystyle \sum$
• May 3rd 2010, 06:13 AM
alternative
Quote:

Originally Posted by mathaddict
hi

This is confusing . You might want to try the latex (a software to format equations) .

Try this ,

logarithms of x to base 3 can be typed as \log_3(x) ,
then wrap it with this $\displaystyle \sum$

http://i669.photobucket.com/albums/v...logaritmik.jpg

Sorry about that, is it better now?

Thanks alot
• May 3rd 2010, 07:00 AM
mrmohamed
Quote:

Originally Posted by alternative
http://i669.photobucket.com/albums/v...logaritmik.jpg

Sorry about that, is it better now?

Thanks alot

HI all
I think x will equal 4 (Itwasntme)
• May 3rd 2010, 07:29 AM
sa-ri-ga-ma
Using change of base (let new base be 2),the given problem can be written as
10log(x)/log(4x) + 21log(x)/log(16x) = 6log(x)/log(x/2)
Cancel log(x) from both side. You get
10/log(4x) + 21/log(16x) = 6/log(x/2)
10/[2 + log(x)] + 21/[4 + log(x) 6/[log(x) - 1]
Let log x to the base 2 is a, then
10/(2+a) + 21/(4+a) = 6/(a-1)
Simplify this equation and solve for a. From that find x.
• May 3rd 2010, 08:35 AM
Soroban
Hello, alternative!

Quote:

$\displaystyle 20\log_{4x}\!\left(x^{\frac{1}{2}}\right) + 7\log_{16x}\!\left(x^3\right) - 3\log_{\frac{x}{2}}\!\left(x^2\right) \;=\;0$

Use the Base-Change formula and change everything to base-2 . . .

. . . . . . $\displaystyle \frac{20\log_2\left(x^{\frac{1}{2}}\right)}{\log_2 (4x)} + \frac{7\log_2(x^3)}{\log_2(16x)} - \frac{3\log_2(x^2)}{\log_2(\frac{x}{2})} \;=\;0$

. . $\displaystyle \frac{20\cdot\frac{1}{2}\log_x(x)}{\log_2(x) + \log_2(4)} + \frac{7\cdot3\log_2(x)}{\log_2(x) + \log_2(16)} - \frac{3\cdot2\log(x)}{\log_2(x) - \log_2(2)} \;=\;0$

. . . . . . $\displaystyle \frac{10\log_2(x)}{\log_2(x)+2} + \frac{21\log_2(x)}{\log_2(x) + 4} - \frac{6\log_2(x)}{\log_2(x)-1} \;=\;0$

Factor: .$\displaystyle \log_2(x)\,\left[\frac{10}{\log_2(x)+2} + \frac{21}{\log_2(x)+4} - \frac{6}{\log_2(x) - 1}\right] \;=\;0$

Multiply through by the LCD. .
I'll drop the "base-2" for now.

. . $\displaystyle \log(x)\,\bigg[10(\log x + 4)(\log x - 1) + 21(\log x + 2)(\log x - 1) - 6(\log x + 2)(\log x + 4)\bigg] \;=\;0$

. . . . . . . $\displaystyle \log(x)\,\bigg[25\log^2x + 15\log x - 130\bigg] \;=\;0$

. . . . . .$\displaystyle 5\log(x)\bigg[\log(x) - 2\bigg]\,\bigg[5\log(x) + 13\bigg] \;=\;0$

And we have three equations to solve:

. . $\displaystyle \log_2(x) \:=\:0\quad\Rightarrow\quad x \:=\:2^0 \quad\Rightarrow\quad \boxed{x \:=\:1}$

. . $\displaystyle \log_2(x) \:=\:2 \quad\Rightarrow\quad x \:=\:2^2 \quad\Rightarrow\quad\boxed{x \:=\:4}$

. . $\displaystyle 5\log_2(x) + 13 \:=\:0 \quad\Rightarrow\quad \log_2(x) \:=\:-\frac{13}{5} \quad\Rightarrow\quad \boxed{x \:=\:2^{-\frac{13}{5}}}$