This page provides a list of educational videos related to Exponential and Logarithmic Equations. You can also use this page to find sample questions, apps, worksheets, lessons , infographics and presentations related to Exponential and Logarithmic Equations.

Ex 2: Solve a Basic Exponential Equation Using the Definition of a Logarithm

By Mathispower4u

Understanding logarithms as inverse exponentials

By Khan Academy

Sal rewrites 100=10^2 as a logarithmic equation and log_5(1/125)=-3 as an exponential equation.

Using logarithms to solve exponential equations

Solving logarithmic equations | Exponential and logarithmic functions | Algebra II | Khan Academy

By Khan Academy

This Khan Academy video on logarithms offers more information on solving logarithmic equations , with a guided example contained in the video as well.

Finding the Inverse of a Function or Showing One Does Not Exist, Ex 1

This course on Second Semester Calculus includes anti-derivatives and integration, then go on to applications of integration areas, volumes, along with inverses exponentials, logarithms , polar, parametric curves, some differential equations and lots of sequences and series. .

Finding the Inverse of a Function or Showing One Does Not Exist, Ex 1

By PatrickJMT

Inverse Relations | MathHelp.com

By MathHelp.com

Here we’re asked to solve for x in the equation: log base x of 144 = 2. Notice that we have a logarithmic equation, so let’s first convert the equation to exponential form. Remember that the base of the log represents the base of the power, the right side of the equation represents the exponent, and the number inside the log represents the result, so we have x…squared…= 144. Now, to solve for x, since x is squared, we simply take the square root of both sides of the equation to get x = plus or minus 12. Remember to always use plus or minus when taking the square root of both sides of an equation. However, notice that x represents the base of the logarithm in the original problem, and the base of a logarithm cannot be negative. Therefore, x cannot be equal to negative 12. So our final answer is x = 12.

Multiplying Scientific Notation | MathHelp.com

By MathHelp.com

In this example, which involves natural logarithms, we’re asked to solve each of the following equations for x, and leave our answers in terms of e. To solve for x in the first equation, ln x = 3, we simply switch the equation from logarithmic to exponential form. Remember that ln x means the natural logarithm of x, and a natural log has a base of e. So, to convert the given equation to exponential form, remember that the base of the log represents the base of the power, the right side of the equation represents the exponent, and the number inside the log represents the result, so we have e…to the 3rd…= x, and we’ve solved for x. Notice that our answer, e cubed, is written in terms of e, which is what the problem asks us to do. Now, let’s take a look at the second equation, ln x squared = 8. Again, we solve for x by switching the equation from logarithmic to exponential form. Ln x squared means the natural logarithm of x squared, and a natural log has a base of e. So, converting the equation to exponential form, we have e…to the 8th…= x squared. Next, since x is squared, we take the square root of both sides. On the right, the square root of x squared is x. On the left, however, there are a couple of things to watch out for. First, remember that the square root of e to the 8th is the same thing as e to the 8th to the ½, which simplifies to e to the 8 times ½, or e to the 4th. Also, remember that when we take the square root of both sides of an equation, we use plus or minus, so our final answer is plus or minus e to the 4th = x.

Solving Logarithmic Equations | MathHelp.com

By MathHelp.com

Here we’re asked to evaluate each of the following logarithms. In part a, we have log base 7 of 49. To evaluate this logarithm, we set it equal to x. In other words, log base 7 of 49 = what? Notice that we now have an equation written in logarithmic form, so let’s see if we can solve the equation by converting it to exponential form. Remember that the base of the log represents the base of the power, the right side of the equation represents the exponent, and the number inside the log represents the result, so we have 7…to the x…= 49. Next, we solve for x. Notice that 7 and 49 have a like base of 7, so we rewrite 49 as 7 squared, and we have 7 to the x = 7 squared, so x must equal 2. In part b, we have log base 3 of 1/27. Again, to evaluate this logarithm, we set it equal to x, and convert the logarithmic equation to exponential form. Remember that the base of the log represents the base of the power, the right side of the equation represents the exponent, and the number inside the log represents the result, so we have 3…to the x…= 1/27. Next, we solve for x. Notice that 3 and 1/27 have a like base of 3, so we rewrite 1/27 as 1 over 3 cubed, and we have 3 to the x = 1 over 3 cubed. Next, 1 over 3 cubed is the same thing as 3 to the negative 3, so we have 3 to the x = 3 to the negative 3, which means that x must equal -3. Therefore, log base 3 of 1/27 = -3. So remember the following rule. To evaluate a logarithm, set it equal to x, convert to exponential form, and solve the equation using like bases.

16 - Simplify Logarithms - Part 1 (Log Bases, Calculate Logarithms & More)

By Math and Science

Quality Math And Science Videos that feature step-by-step example problems!

Evaluating Logarithms | MathHelp.com

By MathHelp.com

In this example, notice that we have a polynomial divided by a binomial, and our binomial is in the form of an x term minus a constant term, or x – c. In this situation, instead of having to use long division, like we did in the previous lesson, we can divide the polynomials using synthetic division, which is a much more efficient method. Here’s how it works. We start by finding the value of c. Since –c = -3, we know that c = 3. Next, we put the value of c inside a box, so we put the 3 inside a box. It’s very important to understand that the number that goes inside the box always uses the opposite sign as the constant term in the binomial. In other words, since the constant term in the binomial is -3, the number that goes inside the box, is positive 3. Next, we write the coefficients of the dividend, which are 2, -7, 4, and 5. Be very careful with your signs. Now, we’re ready to start our synthetic division. First, we bring down the 2. Next, we multiply the 3 in the box times 2 to get 6, and we put the 6 under the -7. Next, we add -7 + 6 to get -1. Next, we multiply the 3 in the box times -1 to get -3, and we put the -3 under the 4. Next, we add 4 + -3 to get 1. Next, we multiply the 3 in the box times 1 to get 3, and we put the 3 under the 5. Finally, we add 5 + 3 to get 8. Now, notice that we have a 2, -1, 1, and 8 in the bottom row of our synthetic division. These values will give us our answer: the first 3 numbers represent the coefficients of the quotient, and the last number is the remainder. And it’s important to understand that our answer will be one degree less than the dividend. In other words, since our dividend starts with x cubed, and we’re dividing by x, our answer will start with x squared. So our answer is 2x squared – 1x + 1 + 8 over x – 3. Notice that we always use descending order of powers in our quotient. In this case x squared, x, and the constant. Finally, remember that we add the remainder over the divisor, just like we did in the previous lesson on long division, and we have our answer. It’s important to understand that we’ll get the same answer whether we use synthetic division or long division. However, synthetic division is much faster.