These rules are guaranteed to solve almost all log problems in competitions, simply by applying laws of logs.
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Take a note of this first though, because this is potentially confusing: When we talk generally about laws of logs, such as log(ab) = log(a) + log(b), we are using the log() function to mean "for any type of log". In other words, this formula means that logx(ab) = logx(a) + logx(b) no matter what x is. This is also true when rewriting a specific log as a fraction of other logs: for example, when we write loga(b) = log(b)/log(a) , we mean that for any value of x, we can say loga(b) = logx(b)/logx(a). When you rewrite a log as a fraction that way, virtually all the time, the logx() functions on the top and bottom will end up cancelling out; for example if we wanted to simplify logab * logba we would write:
So, in a nutshell, when you rewrite a log as a fraction involving logx, the expressions involving x always cancel, which is why we never bother to write x. (In fact, there's a common sense reason why the x's must cancel: the original expression didn't contain any x's, so the final expression definitely can't! If I give you an expression a2bc + b2 and ask you to simplify it and you come up with ab - c + x , then you must have made a mistake somewhere, because you can't start with an expression with no x's in it and end up with an x in the result.) |
The rules:
Rule 1: Replace log(ab) with log(a) + log(b).
If you're working with variables,
this is straightforward:
log(abc) = log(a) + log(b) + log(c).
If you're working with
numbers, however, break the number up into prime factors:
log(600) = log(52) + log(23) + log(3)
(because 600 = 52*23*3). The reason for breaking up the number into
prime factors is because it makes them easier to work with when applying the next rule.
Rule 2: Replace log(a/b) with log(a) - log(b).
Or if you see log(1/b),
replace it with -log(b). That's
a special case of this rule:
log(1/b)
= log(1) - log(b)
= 0 - log(b) (because log(1) is always 0)
= -log(b).
Rule 3: Replace log(an) with n*log(a). Again, if you're working with variables, it's easy. On the other hand, if you're working with numbers, you may need to rewrite the numbers in exponential form. For example if you see log(9) in a problem, it will almost certainly be easier to work with it if you rewrite it as 2*log(3).
Rule 4: Replace logab with log(b)/log(a).
Once you have applied these rules, you should be able to solve the problem using regular algebra. Of course, you can also use regular algebra earlier in the problem as well, if you see a chance to cancel terms and make an expression simpler.
Here is an example that uses all of the rules above:
Simplify: (log3(50) - log3(2))*log53
(log3(50) - log3(2))*log53
| = (log3(52) + log3(2) - log3(2))*log53 | (by applying Rule 1 to log3(50), replacing it with log3(52) + log3(2) ) | ||||||||||||
| = log3(52)*log53 | (by cancelling out log3(2) - log3(2)) | ||||||||||||
| = 2*log35*log53 | (by applying Rule 3) | ||||||||||||
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(by applying Rule 4) | ||||||||||||
| = 2 |