2^x = 8Date: 05/27/2003 at 09:34:52 From: Keith Subject: 2^x = 8 How can I solve for x in the equation 2^x = 8? Date: 05/27/2003 at 09:50:58 From: Doctor Ian Subject: Re: 2^x = 8 Hi Keith, It depends on what you mean by 'solve'. You can start substituting values for x, and see which one works: 2^0 = 1 2^1 = 2 2^2 = 4 2^3 = 8 Success! This doesn't work so well for something like 2^x = 9: 2^0 = 1 2^1 = 2 2^2 = 4 2^3 = 8 Too small 2^4 = 16 Too large At that point, if you have a calculator with a 'y^x' key, you can start trying other exponents: 2^3.1 = 8.57 2^? = 9 It's somewhere between 3.1 and 3.5 2^3.5 = 11.31 Or, you can use logarithms. By definition, if x b = a then log a = x b In this case, if x 2 = 8 then log 8 = x 2 To find a logarithm, you would typically use a calculator or table of logarithms (in much the same way that you'd find the sine or cosine of an angle, and for much the same reason). The fly in the ointment is that most calculators and log tables assume that your base (i.e., 'b' in the description above) is either 10 or e. To deal with other bases (like 2), you need to use the 'change of base' formula, which is explained here: Logarithms: Solving for T http://mathforum.org/library/drmath/view/55561.html Of course, this is kind of a pain to remember if you don't use it on a regular basis, which is why I'd normally use the first method I described, i.e., try exponents until I get a value that I feel is close enough to 9 for my purposes. There is a third approach you can use. If you graph the function f(x) = 2^x - 8 using a graphing calculator, the value of x where the graph crosses the x-axis is the value that satisfies the original equation. Do you see why that's true? I hope this helps. Write back if you'd like to talk more about this, or anything else. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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