Question 26344
ignoring {{{z^x = y^2x}}} for now, lets look at the other 2 equations.


We can take logs to the base2 with {{{2^z = (2)(4^x)}}}. First this is rewritten as {{{2^z = (2)(2^(2x))}}}. This then becomes {{{zlog2 = log2 + 2xlog2}}}


Now log2 to base2 is 1, so we get


{{{z = 1 + 2x}}}
--> 2x - z = -1


x+y+z = 16
x+9+z = 16
--> x + z = 7


Add these 2 equations together and we get 3x = 6. Therefore x = 2.


Hence, from x+z = 7, we then know that z = 5.


So, we have x=2, z=5. Check these values in both of the original equations 2 and 3.


Now check in equation 1... {{{z^x = y^2x}}}. We get {{{5^2 = 9^4}}} ie 25 = 6561 which is clearly wrong. So either you copied one of the equations down wrong? or the answer is "there is no solution to all 3 equations".


jon.