6. Solve for x and check

log5 (x + 3) + log5 (x-1) = 1


We will present two methods of solving this equation. The first involves expressing the left side as a single log as in Problem 5.


log5 (x + 3)(x-1) = 1


Now that we have a single log on the left, we can change to exponential notation as in Problem 3 to get rid of the log.


(x + 3)(x - 1) = 51




x2 + 2x - 3 = 5


We see that we have a quadratic. Transpose all terms to one side leaving a 0 on the other.


x2 + 2x - 8 = 0




(x + 4)(x - 2) = 0


Set the factors = 0


x + 4 = 0 x - 2 = 0




x = -4 x = 2




First let's check the x = 2 in the original equation.


log5 (2 + 3) + log5 (2 - 1) = 1


log5 (5) + log5 (1) = 1


1 + 0 = 1


and it checks.


Now let's check the x = -4


log5 (-4 + 3) + log5 (-4 - 1) = 1


log5 (-1) + log5 (-5) = 1


and we see that we are in trouble. We are trying to compute logarithms of negative numbers. We do not have logarithms defined for negative numbers, so, at this point, the solution does not check.


x = 2


is the only solution.


It is too bad that we haven't defined logarithms of negative numbers yet. If we had them defined in such a way that the rules for logarithms would hold, then we could continue in the check of x = -4 to get


log5 (-1)(-5) = 1


log5 (5) = 1


which checks.


It turns out that there is a way to extend the definition of exponents in order to get the logarithms of negative numbers. However, there is still a problem. It turns out that there are infinitely many complex numbers that will satisfy the definition of a logarithm of any particular negative number, and mathematicians hate it even more when there are several answers than when there are none. For our purposes, we will consider that if you find yourself trying to compute the logarithm of a negative number when you check a solution to a logarithmic equation, then that solution will not check. For this reason, it is necessary to check the solutions to logarithmic equations.


The other method for solving this equation


log5 (x + 3) + log5 (x-1) = 1


would be to put both sides as powers of 5.



Since you add the exponents when you multiply powers of the same base, this is the same as



At this point one of our rules for working with logarithms comes into play. The exponential functions and the logarithmic functions cancel each other out and we get


(x + 3)(x - 1) = 5


which we got with the other method. The same thing that happened with the other method will then happen here and we will get the same result.


This illustrates that changing an equation from logarithmic notation to exponential notation is a consequence of putting both sides as powers on the base.


Return to test