Logarithm cie exam question
- Thread starter Barbie
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This problem requires three bits of experience. The first is the understanding that logs are just numbers and so the normal rules of algebra apply to them when you ignore the special properties of logs. What was one of the first things you were taught in algebra? Gather like terms.
[MATH]log_2(x + 5) = 5 - log_2(x) \implies log_2(x + 5) + log_2(x) = 5 - log_2(x) + log_{2}(x) = 5 + 0 = 5.[/MATH]
This is basic algebra, but you may not see it because you know logs have special properties and so miss the forest for the trees. Logarithms are numbers. Basic algebra still applies. Any questions?
You now can apply the obvious laws of logarithms.
[MATH]log_2(x + 5) + log_2(x) = 5 \implies log_2(\{x + 5\} * x) = 5 \implies log_2(x^2 + 5x) = 5.[/MATH]
This should not cause you to pause if you know this law of logarithms
[MATH]log_a(b) + log_a(c) = log_a(bc).[/MATH]
Still with me?
Now the next bit of experience is that you can easily turn any positive real number into a logarithm of any valid base because of these laws of logarithms:
[MATH]log_a(a) = 1 \text { and } d * log_a(b) = log_a\left ( b^d \right ).[/MATH]
Now the base that is being used on the left of the equation is 2 so that is the base we want in this problem
So we apply this experience to this problem as follows
[MATH]log_2(x^2 + 5x) = 5 = 5 * 1 = 5 * log_2(2) = log_2(2^5) = log_2(32).[/MATH]
There is nothing hard about doing that. What takes a bit of experience is knowing when to do it. Basically, if you have a logarithm equal to a number, do this little procedure to get an equation that equates one logarithm to another in the same base. Why do that?
Because our third and last bit of experience relevant to problems of this type comes from recognizing a very important property of logs that is not always stressed enough in texts or class.
[MATH]log_a(p) = log_a(q) \implies p = q.[/MATH]
[MATH]\therefore log_a(x^2 + 5x) = log_2(32) \implies x^2 + 5x = 32.[/MATH]
So now you should be able to tackle part 2 on your own. But first go back and check that you wrote the problem down correctly because the answer to what you posted is messier than most texts like.
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