Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Solving a Logarithmic Equation - Problem 2

Carl Horowitz
Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Solving a simple logarithmic equation with a single log on one or both sides. In order to solve these equations what we basically want to do is put our log into exponential form. For this particular problem we’re looking at is a logarithm but this time it has a coefficient, there’s two things that we have to do before, one of two things we actually have to do before we actually tackle this problem and both of them include getting rid of this coefficient.

One thing we could do would be to put this coefficient up onto a power, remember our power rule of logarithms so we can take this and put it up, but then what we’re actually going to have to do is we would have to FOIL out this 3x minus 2, not too bad but it could get ugly.

The other way of getting rid of a coefficient would just be to divide it out. This is just a 2 up in front we could divide both sides by 2, it’s gone. I prefer that, I think it’s a little bit easier. If you want to put the 2 up into a power, go right ahead.

I’m going to divide by 2 and this drops out so then we’re just left with log base 4 of 3x minus 2 and that’s just going to equal 3. Now we have a logarithm on one side, no coefficient, all we have to do is put this into exponential form. The 4 comes up and over so we end up with 3x minus 2 is equal to 4 to the 3rd. 4 to the 3rd is 64. This ends up being 3x minus 2 is equal to 64, add 2 to both sides, 3x is equal to 66 and then divide by 3 leaving us with x is equal to 22.

Remember whenever we’re dealing with logarithms of equations, we always have to check our answer. Take our 22, plug it back in here, 3 times 22 is 66 minus 2 is 64 and we just found that the log base 4 of 64 is 3. We just talked about that over here.

So we got rid of our coefficient put it into exponential form, went through our process, checked our work and we actually have an answer in this case, x is equal to 22.

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