Expert: Sherry Wallin - 1/20/2011

Question
Hi,

My finals are coming up soon and there is one problem that throws me off a lot and its mostly knowing where to begin in this problem that I have issues with and if you could help that would be amazing.

The problem is something like this:

LOGbase2[4](LOGbase6[2](LOGbase5[7])))

I have no idea where to begin, so if you can can help would be amazing.

Thank!

Answer
Cam~

You are missing a parenthesis or you have too many parenthesis. Since you said it is a log within a log I am going to assume that you are missing an open parenthesis at the beginning of the problem. There is another problem and you are missing a parenthesis after LOGbase6[2], I believe. So please tell me if this is truly the problem: Incidentally you can use _b to mean base b

(LOG_2[4])[(LOG_6[2])(LOG_5[7])]. This is just multiplication of logs. You do not have logs within logs. What you would need to do is use the change of base formula and make all the logs the same base or change them to their exponential form. (Unless you are fortunate to have a calculator that will take the logs of any base and convert them for you).

The change of base formula says to take the ratio of the log of the argument over the log of the base. i.e., example
log_2(6) = log_b(6)/log_b(2). The argument is 6, the base is 2. I used 'b' for the base because you can convert these logs to any base but generally you use log_10 = log or lne where it is agreed that lne is the natural log base e and is usually just written as ln. I will use base e and the ln function to convert your logs:



[(ln 4)/(ln 2)] [(ln 2)/(ln 6)] [(ln 7)/(ln 5)]. This is easily calculable on your calculator. Notice that if I had used base 10 (log_10 x = log x) on the calculator you will get the same answer:
[(log 4)/(log 2)] [(log 2)/(log 6)] [(log 7)/(log 5)].

Math Prof

PS  Just multiply across the top and across the bottom but notice that these are just fractions or ratios and the log 2's cancel as common factors so you really just have:
[(log 4/1)] [1/(log 6)] [(log 7)/(log 5)]