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Sorry I have to ask such a simple question, my brain is fried after today.

After substituting with a system of equation, I end up with this "simple" logarithmic problem.

$$5 - \log_2 (x - 3) = \log_2(x+1)$$

What property of logs am I looking for to solve this? I know the solution is $x = 7$. Properties of logs looks like it might be my downfall tonight.

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bring the logs to one side, then use $\log a+\log b=\log({a b})$ – David Mitra Dec 20 '11 at 22:07
up vote 9 down vote accepted

Put all log terms on one side: $$5=\log_2(x-3)+\log_2(x+1).$$ Then, write the log terms using only one logarithm (use the law $\log_2 a+\log_2 b=\log_2(ab)$: $$5=\log_2( (x-3)(x+1) ).$$

To get rid of the log, recall, $\log_a x=y\iff a^y=x$, so the above equation can be written as $$2^5=(x-3)(x+1).$$ Can you take it from here?

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I worked through it after your first comment, thanks! – erimar77 Dec 20 '11 at 22:17
@erimar77: Make sure you see why $-5$ is not a solution. – Jonas Meyer Dec 20 '11 at 22:23
Yes, you can't have negative logs when substituting the answer back in. – erimar77 Dec 20 '11 at 22:28
@erimar77: To be precise, you can't have a negative input in a real logarithm. $\log_2 x$ can be negative, like when $x=\frac{1}{4}$, but $x$ can't be negative if $\log_2 x$ is real. – Jonas Meyer Dec 20 '11 at 23:04

Use 5 log2 to the base 2.

as same 5(1)=5log2 to the base 2

You get 25=(x-3)(x+1) as when log with same bases then if log is subtracted to convert logs with same bases the subtracted is divided and if for instance logs are added then to convert then they are multiplied

NOTE: for this the bases have

Hope it helps good luck

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This is largely incomprehensible. To the extent I can understand it it is the same method eloquently displayed in David Mitra's answer. – Jonas Meyer Dec 29 '11 at 19:14

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