 Let us see how we can solve the equation 2 to the x equals negative x plus 11. This is a nice problem where students can break it down into a system of equations, one of which is exponential and the other one which is linear. So let the left hand side of the equation become y equals 2 to the x and the right hand side of the equation let it be y equals negative x plus 11. We can do this now on a graph. We will use a graphing method to solve this because at this stage students probably have not been exposed to solving exponential equations so they do not know how to work with logarithms and so on. So let us use this graphing approach y equals 2 to the x and let the second equation be y equals negative x plus 11. Now let us see how nice. So we see that the functions intersect and we can touch the point where they intersect and we can see that that is the point 3 eighths, that is the point of intersection. When we go back to the system of equations we can verify that solution. The solution was the point 3 eighth and x equals 3 y equals 8. Let us see if that is indeed the case. What is the value of y when x is 3? Well that is indeed 8, 8 is equal to 8. What about here? What is the value of y when x is equal to negative 3? Well we can readily see that also 8 is equal to 8. So you can see that the solution is indeed x equals 3, x is equal to 3 and the function assumes the value of 8.