SOLUTION: write the following in terms of log3, log5 and/or log2. the terms are log27 , log.6, log 5/12, log(base5)6 (I dont really understand what my teacher is asking for and when s

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: write the following in terms of log3, log5 and/or log2. the terms are log27 , log.6, log 5/12, log(base5)6 (I dont really understand what my teacher is asking for and when s      Log On


   



Question 404343: write the following in terms of log3, log5 and/or log2.

the terms are log27 , log.6, log 5/12, log(base5)6
(I dont really understand what my teacher is asking for and when she put and/or log2, does that me that i dont use it on every term? how do i know when to use it or not?)

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
What the instructions mean is:
Rewrite log(27) as an expression of one or more of: log(3), log(5) or log(2).
Rewrite log(0.6) as an expression of one or more of: log(3), log(5) or log(2).
Rewrite log(5/12) as an expression of one or more of: log(3), log(5) or log(2).
Rewrite log%285%2C+%286%29%29 as an expression of one or more of: log(3), log(5) or log(2).

In order to do these problems you have to understand these properties of logarithms:
  • log%28a%2C+%28p%2Aq%29%29+=+log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29
  • log%28a%2C+%28p%2Fq%29%29+=+log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29
  • log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29

One way to see these properties is that they allow you to take a logarithm, the left side, and rewrite it in a different way, with different arguments (the right side). And this is exactly what we are doing in your problems. For example. we are trying to write log(27) in terms of one or more of: log(3), log(5) or log(2).

So the "trick" is to figure out how to express the given argument as a product (the first property), a quotient (the second property), or as a power (the third property, involving zero or more 2's, 3's and 5's. For the first problem we want to express 27 as a product, quotient or power of 2'3, 3's or 5's. There a couple of way to do this one. Since 27+=+3%5E3 we could use the third property:
log%28%2827%29%29+=+log%28%283%5E3%29%29+=+3%2Alog%28%283%29%29
Or we could express 27 as a product, 3*3*3, and use the first property (twice):
log(27) = log(3*3*3) = log(3) + log(3*3) = log(3) + log(3) + log(3) = 3log(3)

For log(0.6) it will help if we rewrite it as a fraction. 0.6 = 6/10 which reduces to 3/5. Now can see that 0.6 is easily expressed as a quotient of a 3 and a 5. So we can use the second property:
log%28%280.6%29%29+=+log%28%283%2F5%29%29+=+log%28%283%29%29+-+log%28%285%29%29

For log(5/12) we already have a quotient and the numerator is already one of the numbers we want to see, 5. All we need to do now is figure out how to express 12 as a product, quotient or power of 2's, 3's or 5's. Fortunately 12 = 2*2*3 so we can use the first and second properties on this one:

Note the use of parentheses above. It is an extremely good habit to use parentheses when substituting one expression for another, especially when the number of terms in the two expressions are different. Here we know that log(2*2*3) = log(2) + log(2*3) (because the first property tells us so)
So when we replace log(2*2*3), a one term expression, with log(2) + log(2*3), a two term expression, we should use parentheses. It helps us know that the subtraction in front applies to both the log(2) and the log(2*3).

For log%285%2C+%286%29%29 we first need to use the change of base formula, log%28a%2C+%28p%29%29+=+log%28b%2C+%28p%29%29%2Flog%28b%2C+%28a%29%29, to change the base of log%285%2C+%286%29%29 to an expression of base 10 logarithms:
log%285%2C+%286%29%29+=+log%28%286%29%29%2Flog%28%285%29%29
Not only have we changed the base to 10, we also now have a log(5) to boot! We now just need to find a way to rewrite log(6). At this point I hope it is easy to see what to do: