Answer:
TWO solutions
Step-by-step explanation:
If I interpret the right half of your equation to be 2^x + 1, and then graph these two functions y = 3x and y = 2^x + 1 on the same set of coordinate axes, the graphs cross (intersect) in two places in Quadrant I.
y = 3x is a straight line graph with slope 3 through the origin (0, 0).
y = 2^x + 1 is an exponential function with y-intercept (0, 2) and horizontal asymptote y = 1.
