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Solving a Logarithmic Equation with Multiple Logs - Problem 3
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Solving a logarithmic equation where we have more than one log on both sides. So whenever we’re dealing with a situation where we have more than one log, what we have to do is use our properties of logs and condense both sides down so we have a single log as equal to a single log.

For this particular problem when we’re looking at is addition on both sides, so that tells us when we condense it we’re actually multiplying. We have the same base, so this just becomes log base 2 of 3x. Doing the same thing one the other side, this just becomes log base 2 of x minus 2 times x plus 2.

Whenever we have a log equal to a log and we have the bases the same we know that the inside have to be the same as well. So we can pretty much just drop our logs leaving us with 3x is equal to x minus 2, x plus 2. We now have our quadratic equation, in order to solve this out, FOIL this out and isolate everything to one side. 3x is equal to x² minus 4, subtract the 3x around and we end up with zero is equal to x² minus 3x minus 4, factor, x minus 4, x plus 1 leaving us with 4 and -1.

Whenever we’re solving logarithmic equations we always, always, always, have to check our answer. Can we plug 4 in and do we get positive numbers inside of our logs? This is always going to be fine because it’s just a number, log base 2 of 4 that works, 4 minus 2 is 2, that works, 4 plus 2 is 6, that works as well.

So we know that our 4 works. Now let’s check the negative one. Again this is just a number but look over here, when we plug in log base 2 of -1, that doesn’t work, we can’t take the log of a negative number so -1 is an extraneous solution and our only solution is 4.

So using the properties of logarithms to condense down logs in order to solve equations.

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