I see no "x in the exponent" here. Or is that "264^15/99xyz" supposed to be "264^(15/99xyz)" rather than "(264^15)/(99xyz)"? What about the "*(2n+1)- (pi+ phi)"? Is that in the exponent or not? Looks like the person who posted that on the internet didn't really know what he was doing. If everything is in the exponent then, like fatra2 said, take logarithms of both sides. If not, subtract the part that is not in the exponent from both sides and then take the logarithm. You don't NEED to take logarithms base 264 (although that gives the simplest result). Logarithms to any base will do.
It probably isn't at all difficult to solve for x, with the exponent interpreted in any way I'd say your problem is finding the log base 264 button on that cash machine.
Haha I would need more than 30 secs just to press all the buttons needed to solve for x, let alone figuring out the answer and what the question actually is - from what it says, I would interpret it as [tex]y=\frac{264^{15}.2xyzn}{99}+1-(\pi+\phi)[/tex] But it could obviously mean many other forms. And like hallsofivy has said, the person is obviously clueless about this kind of stuff and was leaning much towards the humour of this picture rather than the accuracy of its question.
To answer the basic question, "How to solve for x when it's in the exponent?": You would need to isolate the x-term on one side of the equation, take the logarithm of both sides of the equation (any base, as long as they're the same), move x from exponent of a logarithm to coefficient of a logarithm [itex](\log{3^X} = X \log 3)[/tex] Example: [itex]6561 = 3^X[/tex] [itex]\ln{6561} = \ln{3^X}[/tex] [itex]\ln{6561} = X\ln3[/tex] [itex]X = \frac{\ln6561}{\ln3}[/tex] [itex]X = 8[/tex] [itex]\therefore 6561 = 3^8[/tex]