1. Solve The Equation !

Hii !

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

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

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

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

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

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