[MOVED] Find x - y if 2^x^2/2^y^2 = 16^x+y
- Thread starter megan0430
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- Feb 4, 2004
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Do you mean either of the following?
. . . . .\(\displaystyle \L \frac{2^{(x^2)}}{2^{(y^2)}}\,=\,16^{x\,+\,y}\)
. . . . .\(\displaystyle \L \frac{{(2^x)}^2}{{(2^y)}^2}\,=\,16^x\,+\,y\)
Or something else? What have you tried? How far have you gotten?
Thank you.
Eliz.
. . . . .\(\displaystyle \L \frac{2^{(x^2)}}{2^{(y^2)}}\,=\,16^{x\,+\,y}\)
. . . . .\(\displaystyle \L \frac{{(2^x)}^2}{{(2^y)}^2}\,=\,16^x\,+\,y\)
Or something else? What have you tried? How far have you gotten?
Thank you.
Eliz.