For \(x\ \textgreater \ 0\), \(2^x\ \textless \ 3^x\), all the more \(2^x\ \textless \ 3^x+1\), therefore no solution.
For \(x=0\), we have \(2^0=3^0+1 \Rightarrow 1=1+1 \Rightarrow 1=2\). Obviously that' false.
For \(x\ \textless \ 0\) both \(2^x\) and \(3^x\), where \(2^x\ \textgreater \ 3^x\), belong to the range \((0,1)\), then \(3^x+1\) belong to the range \((1,2)\).
Those ranges don't have a common part, therefore, again, no solution.
So, the equation \(2^x=3^x+1\) doesn't have a solution.
