Solving Logarithmic Equations

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En provenance du cours de University of California, Irvine

Pre-Calculus: Functions

368 notes

University of California, Irvine

Pre-Calculus: Functions

368 notes

This course covers mathematical topics in college algebra, with an emphasis on functions. The course is designed to help prepare students to enroll for a first semester course in single variable calculus. Upon completing this course, you will be able to: 1. Solve linear and quadratic equations 2. Solve some classes of rational and radical equations 3. Graph polynomial, rational, piece-wise, exponential and logarithmic functions 4. Find integer roots of polynomial equations 5. Solve exponential and logarithm equations 6. Understand the inverse relations between exponential and logarithm equations 7. Compute values of exponential and logarithm expressions using basic properties

À partir de la leçon

Exponential and Logarithmic Equations.

Exponential and logarithmic functions are two important functions that are useful in applications of calculus. We will discuss properties and graphs of these special functions. We will also learn to solve equations involving exponential and logarithmic functions.

Evaluating Logarithmic Expressions 5:31

Solving Logarithmic Equations 6:02

Exponential Graphs 7:29

Logarithmic Graphs 6:55

Rencontrer les enseignants

Dr. Sarah Eichhorn
Associate Dean, Distance Learning & Tenured Lecturer
Dr. Rachel Cohen Lehman

[SOUND] Let's look at some examples of solve logarithmic equations.

[SOUND] For example, let's solve this equation for x.

We can begin by bringing in the 8 to the right-hand side which would give us that

log base 5 of x + 4 = 9 - 8, or 1. And now writing this in exponential form

gives us that 5^1 = x + 4. Because remember that log base a of x = y

is equivalent to the exponential form a^y = x, which is what we used here.

And now, if we subtract 4 from both sides, we get that x = 5 - 4 or 1.

Now, in solving logarithmic equations, it's very important for us to check our

answers, so let's do that. Let's check that this value, x = 1, satisfies this

equation. Namely, is 8 + log5(1+4) = 9, or is 8 +

log5(5) = 9. And it is because log5(5) is 1, and 8 + 1

is 9, so yes, it works. Therefore, our answer is x = 1.

Alright, let's see another example. [SOUND] Let's solve this equation for x.

Now, we can begin here by condensing the left-hand side into a single logarithm by

using the following property. The logA + logB is equal to the log of

the product, A * B. That is, this equation is equivalent to

log(x-15) * x = 2. And now, we can convert this into

exponential form again by using this equivalence here, and remembering though

that log means log base 10. That is, this equation is equivalent to

10^2 = x-15 * x, or 100 is equal to, and now distributing our x we have x^2 - 15x.

And bringing all the terms to one side gives us that x^2 - 15x - 100 = 0.

And now, let's factor the left-hand side. It factors into (x-20) (x+5).

And now, we have a product of factors equal to zero, which means either the

first factor is zero or the second factor is zero.

When the first factor is zero, that means x = 20, and when the second factor is

zero, that means x = -5. But remember, we have to check these

answers. So, let's start with x = 20.

We'll plug it in here and here. So, the question is,

is log of 20 - 15, which is 5, plus the log of 20 equal to 2?

Again, we can use this property so the question is, is the log of 5 * 20, or

log(100) = 2? And it is, because remember log means log base 10, and 10^2 is indeed

100. So, x = 20 works. It is a solution.

What about x = -5? Well, that would require us to plug in a -5 here, as well

as here. Which we can't do, right? Because log of

-5 - 15, which is log of -20, and log of -5 are both undefined.

Because remember, that the domain of the logarithm is values are strictly greater

than zero. Which means, we're going to cross off x =

-5 over here. And therefore, our only answer is x = 20.

So, it's very important to check your answers when solving logarithmic

equations. Thank you, and we'll see you next time.

[SOUND]

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