Most exponential equations do not solve neatly; there will be no way to convert the bases to being the same, such as the conversion of 4 and 8 into powers of 2. In solving these more-complicated equations, you will have to use logarithms.
Taking logarithms will allow us to take advantage of the log rule that says that powers inside a log can be moved out in front as multipliers. By taking the log of an exponential, we can then move the variable (being in the exponent that's now inside a log) out in front, as a multiplier on the log. In other words, the log rule will let us move the variable back down onto the ground, where we can get our hands on it.
For instance:
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If this equation had asked me to "Solve 2^{x} = 32", then finding the solution would have been easy, because I could have converted the 32 to 2^{5}, set the exponents equal, and solved for "x = 5". But, unlike 32, 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, where I can get at it. And I'll have to use logarithms to bring that variable down.
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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 the variable.
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 (base-10 log, base-2 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 using a base-2 log:
log_{2}(2^{x}) = log_{2}(30)
Any log of the log's base returns a value of 1, so log_{2}(2) = 1. Then:
x · log_{2}(2) = log_{2}(30)
x(1) = log_{2}(30)
x = log_{2}(30)
If you're asked to "find the solution", then the above should be an acceptable answer. However, this value, while "exact", won't be very helpful for word problems (or in "real life") if you need a numerical approximation.
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But we can't evaluate this expression in our calculators as it stands. First, we'd need to apply the change-of-base formula to convert the expression into something in a base that our calculators can understand; namely, the natural log or the common log. That conversion looks like this:
x = log_{2}(30)
Reminder: The "ln" is the abbreviation for "logarithmus naturalis", the German version of what became "natural log" in English. The abbreviation is pronounced "ell-enn" and written with a lower-case "L" followed by a lower-case "N". There is no "I" ("eye") in the function name!
What would happen if I just used the natural log, instead of a base-two log, in the first place? The process would have been exactly the same, and the eventual answer would have been equivalent.
2^{x} = 30
ln(2^{x}) = ln(30)
x · ln(2) = ln(30)
Either way, I get the same answer, but taking natural log in the first place was simpler and shorter.
Note: I could have used the common (base-10) log instead of the natural (that is, the base-e) log, and still come up with the same value (when evaluated in the calculator).
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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 base-5 log of each side, solve, and then apply the change-of-base formula, but I think I'd rather just use the natural log in the first place:
5^{x} = 212
ln(5^{x}) = ln(212)
x · ln(5) = ln(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 base-10 logs can be done on the calculator, I will use the common log instead of the natural log to solve this particular equation:
10^{2x} = 52
log(10^{2x}) = log(52)
2x · log(10) = log(52)
2x(1) = log(52)
2x = log(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:
Since 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:
Then take the log of each side. You'll get an answer in the form:
When you evaluate this, you'll get the same decimal equivalent, 2.866, in your calculator. Don't be shy about being flexible!
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