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Can you multiply two irrational logarithms to get a rational answer?
Yes. Take any rational number p. Let a = any number that is not a power of 10, so that log(a) is irrational. and let b = p/log(a). log(a) is irrational so 1/log(a) must be irrational. That is, both log(a) and log(b) are irrational. But log(a)*log(b) = log(a)*[p/log(a)] = p which is rational. In the above case all logs are to base 10, but any other base can be used.
Log -3 equals?
There is no answer - it is an error: negative numbers do not have logarithms. The log if a number tells to what power the (positive) base must be raised to get the number. Raising any positive number to any power will never result in a negative number, so it is an error to try and take the log of a negative number.
How do you put log functions in a TI-84?
In order to find the log with a power of ten, use the LOG button. For example, to find log105, type log(5). (The parenthesis after the g will appear when you press the LOG button. In order to find a log with a power other than ten, you will have to divide by the log10 of that power. For example, to find log82, type log(8)/log(2). In order to find the natural log of a number, use the LN key. For example, to find the natural log of 91, type ln(91).
Do all logarithms require a base of 10?
No. The so-called "natural" logarithms have a base of ' e ', and you can find the log of any positive number to any base you like.
How do you find the number of years it takes to double a principle on compound interest?
If you have studied logarithms, the answer is simple: The number of years = log(2)/log(1 + r/100) where the annual rate of interest is r%. The logs can be to any base: 10 or e (or any other base). The number of years for it to treble is log(3)/log(1 + r/100) and so on. Without logs, it is a question of trial and error. 100/r year WILL be too large.
