Nov 25 awarded Yearling Nov 25 awarded Yearling Oct 23 awarded Popular Question Sep 30 comment How to solve the equation... Sep 30 comment How to solve the equation...x cannot be 1.. Sep 30 comment How to solve the equation...If $x =1$ , then $y^z = 0$; so $x \neq 1$ since we consider all nonzeroes $x,y$ and $z$. Sep 30 comment How to solve the equation...ok, noted.. wait Ill edit it Sep 30 comment How to solve the equation...thanks @integral, what about all $x, y$ and $z$ are nonzeroes? is there one? Sep 30 revised How to solve the equation...added 47 characters in body Sep 30 comment How to solve the equation...what if $x, y,$ and $z$ are nonzero. Sep 30 comment How to solve the equation...I solved the equation in terms of $x$. but not quite possible. Sep 30 comment How to solve the equation...Sorry, I missed the last closing parenthesis Sep 30 revised How to solve the equation...added 1 characters in body Sep 30 asked How to solve the equation... Sep 27 comment The equation $f(x)= \frac{3^x+1}{2}$ for all positive integers xOk, thanks for that answer.. Sep 27 accepted The equation $f(x)= \frac{3^x+1}{2}$ for all positive integers x Sep 27 comment The equation $f(x)= \frac{3^x+1}{2}$ for all positive integers xisn't it the derivative is $$\frac{f(x+1)-f(x)}{1}$$? so it should be the same, right? so it must be applicable as well? am I right? thanks Sep 27 comment The equation $f(x)= \frac{3^x+1}{2}$ for all positive integers xThanks for the answer, anyway.. why is the derivative of the function not applicable here?my initial approach was to find the derivative, is it wrong? Sep 27 asked The equation $f(x)= \frac{3^x+1}{2}$ for all positive integers x Sep 12 accepted Solving a certain congruence