# Thread: Solve The Equation !

1. ## Solve The Equation !

Hii !

solve in $\displaystyle \mathbb{R}$ the equation :

$\displaystyle \frac{8^x+27^x}{12^x+18^x} = \frac{7}{6}$

2. ## Observation

Note that 8 is 2 cubed, 27 is 3 cubed, 12 is 2 times 6 and 18 is 3 x 6. Also I would cross multiply and note what I just said in the prior sentence to try to work out a solution.

3. Originally Posted by wonderboy1953
Note that 8 is 2 cubed, 27 is 3 cubed, 12 is 2 times 6 and 18 is 3 x 6. Also I would cross multiply and note what I just said in the prior sentence to try to work out a solution.

Its True My Solution ???

4. Originally Posted by Perelman

Its True My Solution ???
I agree with this up to the line $\displaystyle (2^x+3^x)^2 = 25\times6^{x-1}$ and $\displaystyle (2^x-3^x)^2 = 6^{x-1}$. This clearly implies the following line, but is not equivalent to it. In fact, $\displaystyle (2^x+3^x)^2 - (2^x-3^x)^2 = 24\times6^{x-1}$ appears to be a tautology, valid for all x. I certainly don't understand how it implies that $\displaystyle 2\times3^x = 24\times6^{x-1}$.

Starting from $\displaystyle (2^x+3^x)^2 = 25\times6^{x-1}$ and $\displaystyle (2^x-3^x)^2 = 6^{x-1}$ I would take the positive square root on both sides of those equations. The first one gives $\displaystyle 2^x+3^x = 5\times 6^{(x-1)/2}$. For the second equation we need to be more careful. If x>0 then $\displaystyle 3^x>2^x$ and we get $\displaystyle 3^x-2^x = 6^{(x-1)/2}$. Thus the two equations together give $\displaystyle 2^x+3^x = 5(3^x-2^x)$, from which $\displaystyle 3^{x-1}=2^{x-1}$ and therefore x=1.

But if x<0 then the positive square root of $\displaystyle (2^x-3^x)^2$ is $\displaystyle 2^x-3^x$, and a similar procedure to the previous paragraph gives $\displaystyle 2^{x+1} = 3^{x+1}$, so that x=–1.

Thus there are two solutions, x = 1 and x = –1.