You can solve systems of exponential equations algebraically when the bases of the exponential terms are the same number or when obvious (don’t you hate that word in mathematics?) solutions pop out because of the simple natures of the equations involved. If the bases match, you can simply set the exponents equal to each other.
When an algebraic solution isn’t available, then a good graphing calculator or computer program can find the solution — which usually includes lots of decimal values and/or logarithmic functions.
This article deals with the types of problems that you can solve algebraically (or simply). Of course, the most simple is just plugging in a number that you’re pretty sure works. But that method can be time consuming if you have to plug away a lot — reserve it for the sure thing.
Find the common solutions of
and y = 16x+2.
You want the bases to match, so first change the exponential term in the second equation to a power of 2. It becomes y = (24)x + 2 = 24x + 8. Setting the two y-values of the two different equations equal to one another, you get
Now set the two exponents equal to each other: x2 + 6x = 4x + 8. Moving all the terms to the left and factoring, x2 + 2x – 8 = (x + 4)(x – 2) = 0. The solutions of this quadratic equation are x = –4 or x = 2. Replace the x with –4 in either of the original equations, and you get
Replace x with 2 in either equation, and you get y = 65,536.
Find the common solutions of y = 3x + 1 and y = 2x + 3
(0, 3), (–1, 1). A graphing calculator would show you an exponential curve rising from left to right and a line appearing to cut through the curve in two places near the y-axis. You’d have to zoom in closely to see the two points of intersection.
These equations were chosen carefully so that the answers are integers. If you evaluate the two functions for a few values, you can determine the solutions with minimal computation.
Let x = 0 in the first equation, and you get y = 30+1 = 3. Let x = 0 in the second equation, and you get y = 2(0) + 3 = 3. A solution! Let x = –1 in the first equation, and you get y = 3–1+1 = 30 = 1. Let x = –1 in the second equation, and you get y = 2(–1) + 3 = –2 + 3 = 1. These are the only two solutions.
Find the common solution(s) of y = 3x – 1 and y = 9x.
Find the common solution(s) of y = 82 – x and
Find the common solution(s) of y = 2x and y = 1 – x.
Find the common solution(s) of
and y = e.
Following are answers to the practice questions:
The answer is
Set y equal to y to get 3x – 1 = 9x. Change the 9 to 32 and simplify: 3x – 1 = (32)x = 32x. Now that the bases are the same, you can set the two exponents equal to one another and solve for x: x – 1 = 2x; x = –1. Replacing the x with –1 in y = 3x–1, you get
The answer is
First, substitute the exponential expression in the first equation for y in the second. Then change the 8 and 4 to powers of 2, and simplify the equation:
The bases are the same, so set the exponents equal to one another: 6 – 3x = 2x2 – 2x becomes 0 = 2x2 + x – 6. Factoring, you get 0 = (2x – 3)(x + 2). When
and when x = –2, y = 82–(–2) = 84 = 4,096.
The answer is (0, 1).
The first equation is an exponential that rises steadily as the x-values increase. The graph of the second equation is a line that falls steadily from left to right. They intersect at a single point. With some careful selections of points, you can quickly determine their single common solution, (0, 1).
Replacing the x with 0 in the exponential gives you y = 20 = 1. And replacing the x with 0 in the line gives you y = 1 – 0 = 1.
The answer is (1, e), (–1, e).
The exponential function is positive for all values of x that you input. And the line is horizontal with a y-intercept of (0, e). If you replace the x in the exponential with 1, you get y = e1 or y = e. The same thing occurs when you replace the x with –1; the square of –1 is also 1, so you get the same y-value.