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logarithm question

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1 Answer

Daniel,
 
Let me make sure I understand the question correctly:  You need to write the following logarithm as a sum, difference, or product of logs...
 
log8  (x2)/(y*z2)
 
Before I get to answering your question, let me review the log rules quickly:
  1. Product rule:  log (a*b) = log (a) + log (b)
  2. Division rule:  log (a/b) = log (a) - log (b) (ORDER MATTERS!)
  3. Exponent rule:  lob (a2) = log (a)2 = 2*log (a)  (This is NOT the same as (log a)2!!!)

Now that we have those rules, let's apply them to this question.  Start with the math operation that is connecting all parts of (x2)/(y*z2) together: the division sign.

Apply Rule #2:

log8 (x2)/(y*z2)   -->  log8 (x2) - log8 (y*z2)

Now we have two separate logs to work with.  Let's focus on log8 (x2) first.

Apply Rule #3:

log8 (x2)   -->   2*log8 (x)

Now for the second part: log8 (y*z2).

Apply Rule #1:

log8 (y*z2)   -->   log8 (y) + log8 (z2)

If you notice, this part can be simplified further.  Apply Rule #3 to the log containing z2.

log8 (y*z2) --> log8 (y) + log8 (z2)   -->   log8 (y) + 2*log8 (z)

Now that everything has been simplified down as far as we can go, let's update the whole expression:

log8 (x2)/(y*z2) --> log8 (x2) - log8 (y*z2)   -->   2*log8 (x) - [log8 (y) + 2*log8 (z)]

If you're comparing this answer to the answers in the back of a textbook, they may take this one step further and distribute the subtraction sign through to the log8(y) and the 2log8(z), in which case, your answer would look like this:

2*log8 (x) - log8 (y) - 2*log8 (z)

Hope this is helpful!!

~Patty