Evaluating Logarithms | MathHelp.com - Free Educational videos for Students in K-12 | Lumos Learning

Evaluating Logarithms | MathHelp.com - Free Educational videos for Students in k-12


Evaluating Logarithms | MathHelp.com - By MathHelp.com



Transcript
00:0-1 here were asked to evaluate each of the following logarithms
00:05 . In part a we have log based seven of
00:08 49 . To evaluate this law algorithm , we set
00:13 it equal to X . In other words log based
00:17 seven of 49 equals what Notice that we now have
00:23 an equation written in log arrhythmic form . So let's
00:27 see if we can solve the equation by converting it
00:31 to exponential form . Remember that the base of the
00:35 log represents the base of the power . The right
00:39 side of the equation represents the exponent and the number
00:43 inside the log represents the result . So we have
00:48 seven to the X equals 49 . Next we solve
00:55 for X . Notice that seven and 49 have a
00:59 like base of seven . So we rewrite 49 as
01:04 seven squared and we have seven to the X equals
01:10 seven squared . So X must equal to in part
01:15 B . We have log base three of one 27th
01:21 . Again to evaluate this law algorithm , we set
01:24 it equal to X . and convert the law algorithmic
01:28 equation to exponential form . Remember that the base of
01:33 the log represents the base of the power , the
01:36 right side of the equation represents the exponent and the
01:40 number inside the log represents the result . So we
01:45 have three to the X Equals 1/27 . Next we
01:52 solve for X . Notice that three and 1/27 have
01:57 a like base of three . So we rewrite one
02:01 27th as 1/3 cubed and we have three to the
02:08 X equals 1/3 cubed . Next 1/3 cubed is the
02:15 same thing as three to the negative three . So
02:20 we have three to the X equals three to the
02:23 negative three , Which means that X must equal negative
02:28 three . So remember the following rule to evaluate algorithm
02:34 set it equal to X , convert to exponential form
02:39 and solve the equation using like bases .
Summarizer

DESCRIPTION:

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.

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