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Solving Exponential
Equations: Sections: Solving from the definition, Solving using logarithms, Calculators Most exponential equations do not solve neatly; in solving these morecomplicated equations, you will need to use logarithms. For instance:
If this equation had asked me to "Solve 2^{x} = 32", it would have been easy, because I could have converted the 32 to 2^{5}, set the exponents equal, and solved for "x = 5". But 30 is not a power of 2, so I can't set powers equal to each other. I need some other method of getting at the x, because I can't solve with the equation with the variable floating up there above the 2; I need it back down on the ground where it belongs. And I'll have to use logarithms to get at it. When dealing with equations, I can do whatever I like to the equation, as long as I do the same thing to both sides. And, to solve an equation, I have to get the variable by itself on one side of the "equals" sign; to isolate the variable, I have to "undo" whatever has been done to it. In this case, the variable x has been put in the exponent. The backwards (technically, the "inverse") of exponentials are logarithms, so I'll need to undo the exponent by taking the log of both sides of the equation. This is useful to me because of the log rule that says that exponents inside a log can be turned into multipliers in front of the log: log_{b}(m^{n}) = n · log_{b}(m) When I take the log of both sides of an equation, I can use any log I like (base10 log, base2 log, natural log, etc), but some are sometimes more useful than others. Since the base in the equation "2^{x} = 30" is "2", I might try logbase2: log_{2}(2^{x})
= log_{2}(30)
But I can't evaluate this in my calculator unless I apply the changeofbase formula: x
= log_{2}(30) What would happen if I just used the natural log in the first place? 2^{x}
= 30
Either way, I get the same answer, but taking natural log in the first place was simpler and shorter. Copyright © Elizabeth Stapel 20022011 All Rights Reserved Note: I could have used the common (base10) log instead of the natural (basee) log, and still come up with the same value (when evaluated in the calculator). Since science uses the natural log so much, and since it is one of the two logs that calculators can evaluate, I tend to take the natural log of both sides when solving exponential equations. This is not (generally) required, but is often more useful than other options.
Since 212 is not a power of 5, then I will have to use logs to solve this equation. I could take base5 log of each side, solve, and then apply the changeofbase formula, but I think I'd rather just use the natural log in the first place: 5^{x}
= 212
...or about 3.328, rounded to three decimal places.
Since 52 is not a power of 10, I will have to use logs to solve this. In this particular instance, since the base is 10 and since base10 logs can be done on the calculator, I will use the common log instead of the natural log to solve this equation: 10^{2x}
= 52
...or about 0.858, rounded to three decimal places.
Before I can start looking at the exponential, I first have to get rid of the 3, so I'll divide that off to get: 2^{x+4} = ^{350}/_{3} Since ^{350}/_{3} is not a power of 2, I will have to use logs. I will use the natural log in this case: ...or about 2.866, rounded to three decimal places. Note: You could also solve the above by using exponent rules to break apart the power on the 2: 2^{x}^{+4} = (2^{x})(2^{4}) = (2^{x})(16) Then divide through by the 16 and simplify to get 175/24. Then take the log of each side. You'll get an answer in the form ln(175/24)/ln(2). When you evaluate this, you'll get the same decimal equivalent, 2.866, in your calculator. Don't be shy about being flexible! << Previous Top  1  2  3  Return to Index Next >>



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