Algebra Help: Solving Exponential Equations

Solve exponential equations with algebra.
Solve exponential equations with algebra.

Though they may look more intimidating than simple linear equations, exponential equations can be solved using the simple algebraic principle of applying the same operations to both sides of the equation. To solve for an unknown exponent, you need to use logarithms, which are inverses of exponential functions. Think of applying logarithms to both sides like pulling the strings of your shoelaces, pull them all the way out and the knot becomes undone!

This guide will show you how to solve an exponential equation using the example problem given on the right. In case you can't see the image, it's

7.45(1.09)^x = 1.4(2.113)^x

There are also a few more examples worked out.

Step 1

Let's consider the sample problem 7.45(1.09)^x = 1.4(2.113)^x. The first step is to take the logarithm of both sides of the equation. It does not matter whether you use the natural logarithm function ln(x), or the base-10 logarithm function log(x). In this example we will use the function ln(x). Thus, the original expression becomes

ln[7.45(1.09^x)] = ln[1.4(2.113^x)]

Step 2

Simplify the logarithm expression using the two basic rules of logs:

log(a*b) = log(a) + log(b)

log(c^d) = d*log(c)

This gives us

ln[7.45] + x*ln[1.09] = ln[1.4] + x*ln[2.113]

which is now a linear algebraic equation in the variable x.

Step 3

Combine like terms, moving the x terms to one side of the equation and the constant terms on the other.

ln[7.45] - ln[1.4] = x*ln[2.113] - x*ln[1.09]

ln[7.45] - ln[1.4] = x*(ln[2.113] - ln[1.09])

Step 4

Convert the log values into numbers and solve for x. You can use a scientific calculator to evaluate the natural log function, or tables of logarithms. For the most accurate answer, use a calculator.

1.67174 = x*0.66193

1.67174/0.66193 = x

2.52555 = x

To check your work, you can use the exponential equation solver which solves equations of the form ab^x = cd^x. Just input the values of a, b, c, and d. In this example, a = 7.45, b = 1.09, c = 1.4, and d = 2.113.

Solving Equations (Source: Edu-Stock)
Solving Equations (Source: Edu-Stock)

Another Example

Solve the rational-exponential equation

[2(3^x) + 1]/[3^x - 16] = 5

First multiply both sides of the equation by the denominator 3^x - 16. This creates the simpler and equivalent equation

2(3^x) + 1 = 5(3^x - 16)

2(3^x) + 1 = 5(3^x) - 80

It's important to resist the temptation to incorrectly simplify the expression. 2(3^x) does not equal 6^x, and 5(3^x) does not equal 15^x. Instead, you must treat 2(3^x) and 5(3^x) as like terms and combine them on one side of the equation.

2(3^x) - 5(3^x) = -80 - 1

(2 - 5)(3^x) = -81

-3(3^x) = -81

3^x = 27

At this point you can either use logarithms to solve for x, or you can use basic math skills to deduce that x = 3, since 3^3 = 3*3*3 = 27.

Last Example

A substance has a half-life of 4 hours. If you start with 9 grams, how long will it be until you have 3 grams? To solve this equation, we start with the expression

Q(t) = 9*e^(rt)

where Q(t) is the quantity of the substance after t hours, e is the mathematical constant 2.71828..., and r is an unknown rate factor. Since the half-life is 4 hours, we know that if t = 4 then Q(t) will equal 4.5. This let's us solve for r:

4.5 = 9*e^(r*4)

4.5/9 = e^(4r)

0.5 = e^(4r)

Ln(0.5) = 4r

Ln(0.5)/4 = r

So the value of r is approximately -0.1732868. Using this we can now solve the original question, find the time t when Q(t) equals 3 grams.

3 = 9*e^(-0.1732868t)

3/9 = e^(-0.1732868t)

1/3 = e^(-0.1732868t)

Ln(1/3) = -0.1732868t

Ln(1/3)/-0.1732868 = t

So t = 6.33985 hours.

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billybuc 23 months ago from Olympia, WA

Oddly, at the high school I attended, we had a choice of taking the math track or the science track. Since science was a mystery to me, I chose math. I always liked algebra because it was a puzzle to solve. These secrets would have come in handy back then. :)

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