Solving this Exponential Equation This is the equation that I need help with. The fact that there is that extra $1$ is throwing me off. If you move the $4^x$ term over and take the $\log$ of both sides, then you have a $\log$ with a polynomial inside.
$$5^x - 4^x = 1$$

The function $y(x)=5^x-4^x$ is negative if $x<0$ . So $x<0$ is excluded. For $x>0$  the function is strictly increasing from $0$ to $\infty$. So, there is only one value of $x$ so that $y=1$. 
Obviously this value is $x=1$ because  $y(1)=5^1-4^1=1$
