If the dependent variable of an exponential function written in the form y=a⋅bx, is exchanged for a constant, say C, the result is a one-variable equation: C=a⋅bx.
This type of equation is called an exponential equation, and can be solved graphically. This is done by first graphing the function y=a⋅bx, then finding the x-coordinate of the point(s) on the graph with the y-coordinate C. The x-coordinate(s) is the solution to the equation.Use the graph to solve the equation 3=5⋅0.85x.
The graph shows all x-y points that satisfy the function rule y=5⋅0.85x. Let's compare the function rule and the equation. Function rule:y=5⋅0.85xEquation:3=5⋅0.85x The only difference between these two equalities is that the independent variable, y, is replaced by a 3 in the equation. Thus, we solve the equation by finding the x-coordinate of any point on the graph that has the y-coordinate 3.
We can identify one such point in the graph. Let's now find the x-coordinate of this point graphically.
This x-coordinate is not easily read from the graph, so we'll have to make an approximation. It's just a bit bigger than 3, so we'll use 3.1. This means that an approximate solution to the equation is x≈3.1. We can verify this by substituting it into equation to see if a true statement is made.
The right-hand side and the left-hand side are approximately equal, so we have indeed found an approximate solution to the equation: x≈3.1.
There are different ways to algebraically solve exponential equations. If both sides of the equation can be written in the same base, equality can be used. For example, consider the equation 42x=642. Since 64=43, the equation can be written as follows. 42x42x=(43)2=46 Now, we have two equivalent expressions with the same base. For the equality to hold, the exponents must also be equal. Thus,
2x=6⇔x=3.Solve the equation 35x−3=3-2x+11.