 ###### 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 1

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. What I mean by simple logarithmic equation is basically we have a single log on each side or a log and a number. So in this problem what’s inside of our logarithm isn’t particularly simple. We have a quadratic inside but we still only have a single log. So what we can do is put this into exponential form and then get rid of our logarithm.

So putting this into exponential form, we end up with x² plus 10x plus 25 is equal to 3 to the 4th. 3 to the 4th is 81, so this ends up with x² plus 10x plus 25 is equal to 81. From here we have a quadratic equation. We bring everything to one side, subtract the 81 over, x² plus 10x, 25 minus 81, just make sure we get it right, so we go to our calculator, I think it’s 56, yeah, minus 56 is equal to zero. Now we have to factor.

So we end up with x and x, one has to be minus one has to be plus, because our last term was negative and we need our positive term to be bigger. Factors of 56 are 1 and 56, that’s obviously not going to work. 7 and 8, again that’s not going to give us 10 and then 4 goes into 56 as well, we end up with 4 times 14, which will end up giving us 10. If this is minus 4 and plus 14 we’re going to work out to be 56 and positive 10. Finding our solutions we know that x can be equal to 4 or -14.

Now we always have to be careful with our domain when we’re solving this, we can’t take the logarithm of a negative number, so we have to go back and plug in these answers to make sure it actually works with our original equation. Take 4, 4² is 16, plus 40 is 56 plus 25, we already did that here is 81. Log to base 3 of 81 is 4. So we know that 4 actually works. -14, these numbers are going to be a little bit bigger, so I’m going to use my calculator, - 14², when you square a negative, works out to be positive so this is 196, minus 140, which is -14 times 10, is 56 and again plus 25 will leave us with 81. Negative 14 works as well. We actually end up with two solutions to this logarithmic equation.

One thing to be careful of is whenever we’re solving logarithmic equations, we can have a negative number as our answer, we just have to make sure that when we plug it into our logarithm, it actually ends up being positive. So whenever we plug our number in here, this has to be positive, it doesn’t matter what our answer is as long as that holds true.