A number theory problem by Hemanth K

For $\log_2 n$ to be an integer, $n$ must be a power of 2. Similarly, For $\log_3 n$ to be an integer, $n$ must be a power of 3. The only power possible is 0, that is $2^0 = 3^0 = 1$ and $\log_2 1 = \log_3 1 = 0$ , therefore, there is only $\boxed{1}$ integer $n$ satisfies the conditions.

means 2^x=3^y; it is only one solution that is x=0,y=0 means n=1; 2^x is even and 3^y is odd so they both are not equal so no other answer