Exponential equation

Erzenwald · #1 $Old$ November 9th, 2008, 05:11 AM

I'm having trouble solving the following equation

$3*4^x - 6^x = 2*9^x$

I managed to get it to a simpler form but I'm stuck there

I would appreciate any help.

mr fantastic · #2 $Old$ November 9th, 2008, 05:38 AM

Quote:

Originally Posted by Erzenwald $View Post$

I'm having trouble solving the following equation

$3*4^x - 6^x = 2*9^x$

I managed to get it to a simpler form but I'm stuck there

I would appreciate any help.

I doubt an exact solution can be found. Is there any reason why an approximate solution can't be given?

Rapha · #3 $Old$ November 9th, 2008, 06:02 AM

Quote:

Originally Posted by mr fantastic $View Post$

I doubt an exact solution can be found. Is there any reason why an approximate solution can't be given?

I think, because the trivial solution is just x = 0 in IR

proof:

3*4^0 - 6^0 = 2*9^0

3*1-1 = 2*1

2 = 2

true

Erzenwald · #4 $Old$ November 9th, 2008, 06:42 AM

I found a solution

$3*4^x - 6^x = 2*9^x$ We divide all by $2*9^x$

$\frac{3}{2}*\frac{{4^x }}{{9^x }} - \frac{1}{2}*\frac{{6^x }}{{9^x }} = 1$
We get the equation to a simpler form

$\frac{3}{2}*\frac{{2^{2x} }}{{3^{2x} }} - \frac{1}{2}*\frac{{3^x 2^x }}{{3^{2x} }} = 1$

$\frac{3}{2}*\frac{{2^{2x} }}{{3^{2x} }} - \frac{1}{2}*\frac{{2^x }}{{3^x }} = 1$

Now we can write

$\frac{{2^x }}{{3^x }}$ as Y and solve the quadratic equation