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

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- May 3rd 2010, 05:18 AMalternativeSolving 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 AMmathaddict
- May 3rd 2010, 06:13 AMalternative

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

Sorry about that, is it better now?

Thanks alot - May 3rd 2010, 07:00 AMmrmohamed
- May 3rd 2010, 07:29 AMsa-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 AMSoroban
Hello, alternative!

Quote:

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

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Factor: .

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

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And we have three equations to solve:

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