Evaluating Logarithms | MathHelp.com - By MathHelp.com
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 . |
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 â 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. Hereâs how it works. We start by finding the value of c. Since âc = -3, we know that c = 3. Next, we put the value of c inside a box, so we put the 3 inside a box. Itâs 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, weâre 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 itâs 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 weâre dividing by x, our answer will start with x squared. So our answer is 2x squared â 1x + 1 + 8 over x â 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. Itâs important to understand that weâll get the same answer whether we use synthetic division or long division. However, synthetic division is much faster.
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