Composite Functions Using LogarithmsDate: 3/10/96 at 21:33:41 From: Anonymous Subject: Precal I have the following precal question and am having a bit of difficulty with it. Any help (even an idea about how to approach it) would be helpful. Thanks so much. Suppose f and g are functions defined by f(x) = x+2 and g(x) = x. Find all x > -2 for which g(x) * logbase3 f(x) 3 = f(x) In case you can't read that, it says 3 to the exponent (g(x) * the log f(x) base3) equals f(x). Thanks. Date: 3/21/96 at 13:32:36 From: Doctor Aaron Subject: Re: Precal Hello, Good question. A good place to start is to see that you have an expression that is close to the form: logbasea (f(x)) a . If this doesn't look useful, think about what the log function does. It is essentially the inverse of the exponential function (provided the exponential and the log have the same base). Composing a function with its inverse gives us what is called the identity function - it gives us back whatever we give it to operate on. That is, if we raise a number to a power, and then take the log (of the appropriate base) we get back the power to which we raised our number. Similarly, if we take the log of a number, then raise the base to the power (log(base)), we get our number back. so then logbasea (f(x)) a = f(x). You may have noticed a problem. You have g(x) in front of the log, so we can't simplify as nicely as we would like to. We can change this by using the relation that a*log(b) = log(b^a) where ^ is an exponent. This relation makes sense, because the log just asks us how many times we have to multiply a number by itself to get another number. (Actually we can't multiply something by itself a fractional number of times, but log is just the extension of the above definition to the real numbers) Well, if we have to muliply (base) by itself n times to get b, we'll have to multiply (base) by itself a*n times to get b^a. Now rewrite g(x) * logbase3 f(x) as logbase3 (f(x)^g(x)). You should be able to use the above ideas to make some headway. If you run into some more trouble, consult a teacher or write back. -Doctor Aaron, The Math Forum |
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