Find A and B (defined by x and y)

(2^x*Square root3^y)/(2^(y-x)* 1/3^x) = 2^A * 3^B

Question

Find A and B (defined by x and y)

(2^x*Square root3^y)/(2^(y-x)* 1/3^x) = 2^A * 3^B

anonymous · Answer

Is that 1 over 3^x?

anonymous · Answer

Yea

anonymous · Answer

So it is actually  2 to the power[ (y-x) / 3^x], right?

anonymous · Answer

It's $$2^{x} * \frac{ 1 }{ 3^{x} }$$

anonymous · Answer

The second part, not the first....

anonymous · Answer

(2^ (x*Square root(3^y))) / (2^((y-x)* 1/3^x))

Is this right?

anonymous · Answer

$$\frac{ 2^{x}*\sqrt{3^{y}} }{ 2^{y-x} *\frac{ 1 }{ 3^{x} } } = 2^{A} * 3^{B}$$

Thats the whole

anonymous · Answer

If so, since everything is to power 2 (top and bottom) you can just transfer the denominator to the numerator so

2 ^ (x * cubert (3y) - 3 power -x * (y-x))  and then just compare to find A and B

anonymous · Answer

That wont be right, seeing as you used 2 ^(x*...), or will it?

anonymous · Answer

Ah, I didn't see that you changed the equation......

anonymous · Answer

I had ^(x*cuberoot y) and you put  (^x) * cuberoot y

But still, the principle is same , bring everything up to the top

eg 2^x and 2^(y-x)  just become 2^y on top etc

anonymous · Answer

Ah, i start to see it now.

anonymous · Answer

Great:-)