Rules Logarithm Worksheet Solutions
NAME:
Remember to think of log b x as the number to which I raise b to get x. This is very important in the study of logs. Let b, v, and w be positive real numbers where b is not equal to one. Let k be a real number. 1. In words, what is log b b ? It's the number to which I raise __b___ to get __b___. What does this number we call log b b have to be? Any number raised to 1 is itself. The number you raise b to, to get itself, has to be 1. So log b b = 1 . 2. In words, what is log b 1 ? It's the number to which I raise __b___ to get __1___. What does this number we call log b 1 have to be? Any number raised to 0 is 1. The number you raise b to, to get 1, has to be 0. So log b 1 = 0 . 3. In words, what is log b (b k ) ? It's the number to which I raise __b___ to get __bk___. What does this number we call log b (b k ) have to be? This is sort of a trick question. What do I raise 2 to, to get 23 ? (Well, 3, silly. Don't ask stupid questions.) So, what do we raise b to, to get bk? (Well, k, silly. I told you not to ask stupid questions.) So log b (b k ) = k . 4. Now log b v is the number to which I raise b to get v. Therefore b raised to this power log b v or should be what number?
b
We have to hold in our head that log b v is the number to which I raise b to get v. This is crucial. See this " log b v " as one entity. It's the number I log b v raise b to, to get v. Now, when we write , we have taken b and log b v raised it this number. I should get v. So . We can also see log1 0 1000 this be looking at specific examples. We see that 10 = 10 3 = 1000 .
b b
=v
5. Complete the table. log 3 3 = 1 log 5 25 = 2 log 3 9 = log 5 5 = 2 1 2 log 3 27 = log 5 125 = 3 3 6
log 2 16 = 4
log 2 4 =
log 2 64 =
Now use the three rows of the table to determine the relationship between log b v , log b w , and log b (vw) . Write the rule down and then show it works using an example from the table. In the first row of the table, notice how we have the log of 3, the log of 9, and the log of their product 27. The second row finds the log of 25, the log of 5, and the log of their product 125. This is true of the third row too. What is the connection between (first row answers) 1 and 2, and 3? What is the connection between (third row answers) 4 and 2, and 6? We notice if we add the first two columns, we get the third column. So we see the rule emerge: log b (v ) + log b ( w) = log b ( vw) . As an example, log 2 16 + log 2 4 = log 2 64 . This is one of many the rules we will use to manipulate logs. 6. Complete the table. log 3 27 = 3 log 5 125 = 3 log 3 9 = 2 log 5 25 = 2 2 log 3 3 = 1 log 5 5 = 1 4
log 2 64 =
6
log 2 4 =
log 2 16 =
Now use the three rows of the table to determine the relationship between log b v , v log b w , and log b ( ) . Write the rule down and then show it works using an example w from the table. The reasoning is similar to that used above. The rule we are after is v log b (v ) - log b (w) = log b ( ) . As an example, log 2 64 - log 2 4 = log 2 16 . w
7. Complete the table. You may use your calculator for the first two. Round your answers to three decimal places. The last one you should be able to work out logically. log 35 =
log e 6 2 = log 5 5 2 =
( )
2.386 3.584 2
5 * log 3 =
2.386 3.584 2
( ) ( )
2 * log e 6 = 2 * log 5 5 =
Now use the three rows of the table to determine the relationship between log b v k and k log b v . Write the rule down and then show it works using an example from the table. To find log 5 (5 2 ) by hand, use the rule developed in #3. To find 2 * log 5 5 , find log 5 5 and then multiply by 2. You should see that, for each example above, log b v k = k log b v . As an example, log 5 (5 2 ) = 2 * log 5 5 .
Additional note concerning rules log b (b k ) = k and
b log b v = v :
These two rules are true because of the inverse relationship between the exponential function f ( x ) = b x and the logarithmic function g ( x) = log b ( x ) . Recall inverse functions undo each other....