Logarithmic and Exponential Equations
Date: 11/28/2006 at 17:35:43 From: Patrick Subject: solve 1/9 = 27^(2x) My son is taking Algebra II. He is a very bright student, but he does not understand exponential and logarithmic equations. The above is an example of just one question. The entire equation is confusing. He does not "get it". Can you please help?
Date: 11/29/2006 at 10:42:57 From: Doctor Ian Subject: Re: solve 1/9 = 27^(2x) Hi Patrick, Logarithms are just a different way of thinking about exponents, in the same sort of way that division is a different way of thinking about multiplication. For example, 3 * 4 = 12 3 = 12 / 4 4 = 12 / 3 are just three different ways of saying the same thing. One form is nice when you know the factors, 3 * 4 = ? and the other is nice when you know the product and one factor, 12 / 3 = ? but if given any form, you can write it as one of the others without changing the meaning: 12 / 3 = ? ---------> 12 = 3 * ? means the same as And in many cases, changing form can help you see what's going on, or just do a computation more easily. In much the same way, 2^3 = 8 log_2(8) = 3 are just two different ways of saying the same thing, and you can change either form to the other. If you play around with this relationship enough to get comfortable with it (and, as with learning to hit a baseball, there's no other way to do it, except get some practice), then you'll see that an exponent and a logarithm are inverses: 2^[log_2(8)] = 8 The log inverts, or 'cancels', \______/ the exponent I've just replaced 3 with log_2(8), which it equals. log_2[2^3] = 3 Same deal. \_/ I've just replaced 8 with 2^3, which it equals. Does this make sense? If so, let's move on. (If not, let me know so we can talk about this some more.) Let's look at your equation: 1/9 = 27^(2x) The first thing you'd like to do is find a common base for all the expressions, because that lets you use the various rules for manipulating exponents: Properties of Exponents http://mathforum.org/library/drmath/view/57293.html In the example, everything is a power of 3, so we want to express everything as '3 to the something': 1/(3^2) = (3^3)^(2x) Now we can use those rules I mentioned earlier. We can write the left side as 3^(-2) = (3^3)^(2x) and the right side as 3^(-2) = 3^(3*2x) = 3^(6x) Now there are a couple of ways we can think about this. One is to just realize that if a^b = a^c it has to be the case that b = c But we can also use logarithms to get rid of the bases: 3^(-2) = 3^(6x) log_3[3^(-2)] = log_3[3^(6x)] -2 = 6x And now we're home-free. But here's a very important thing to note: In the final step, we're going to invert the operation of multiplication, by dividing: -2 6x ---- = ---- Dividing by 6 inverts, or 'cancels', 6 6 the multiplication by 6 on the right. -2/6 = x This is EXACTLY THE SAME IDEA as using a logarithm to invert, or cancel, an exponent; or using an exponent to invert, or cancel, a logarithm. It's just that the form looks a little funny, in that we use a functional notation, rather than a new operator. It might help to think of all this in terms of the 'bigger picture'. Once we've defined an operation like addition, 3 + 2 = 5 we can think about a situation where we are missing one of the operands, 3 + ? = 5 and we can create a new operation (subtraction) that inverts addition: 3 + ? = 5 ---> ? = 5 - 3 Later, when we've defined an operation like multiplication, 3 * 2 = 6 we can think about a situation where we are missing one of the operands, 3 * ? = 6 and we can create a new operation (division) that inverts multiplication: 3 * ? = 6 ---> ? = 6 / 3 Still later, when we've defined an operation like exponentiation, 2^3 = 8 we can think about a situation where we are missing one of the operands, 2^? = 8 ?^3 = 8 and we can create a new operation (logarithms) that inverts exponentiation: 2^? = 8 ---> log_2(8) = ? So this is one of the ways that math grows: We figure out something we can do, and then we ask: Is there a way to UNDO it, if we know what comes out, but not what went in? And then we have something ELSE that we can do, i.e., we get two games for the price of one. :^D Let me know if this helps. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum