# Question Video: Finding the Solution Set of Logarithmic Equations over the Set of Real Numbers Mathematics • 10th Grade

Determine the solution set of the equation logβ (π₯Β² β 5π₯ + 4) = 4 + logβ (π₯ β 1) in β.

02:20

### Video Transcript

Determine the solution set of the equation log base three of π₯ squared minus five π₯ plus four is equal to four plus log base three of π₯ minus one in the set of all real numbers.

We begin by collecting all the logarithmic terms on one side of our equation. We can do this by subtracting log base three of π₯ minus one from both sides. This gives us the equation as shown. We can then use one of our laws of logarithms on the left-hand side. Log base π of π₯ minus log base π of π¦ is equal to log base π of π₯ divided by π¦. It is important to note that this rule only holds if our terms have the same base. Our equation therefore becomes log base three of π₯ squared minus five π₯ plus four divided by π₯ minus one is equal to four.

The numerator of the expression weβre taking the logarithm of can be factored. π₯ squared minus five π₯ plus four is equal to π₯ minus four multiplied by π₯ minus one. This is because negative four and negative one have a sum of negative five and a product of four. We notice that the numerator and denominator both contain π₯ minus one. This means that these terms cancel as long as π₯ is not equal to one. And as weβre given log of π₯ minus one in the question, which is not defined when π₯ is one, then π₯ cannot equal one. Log base three of π₯ minus four is equal to four.

We can now use another one of our laws of logarithms. If log base π of π₯ is equal to π, then π₯ is equal to π to the power of π. We can therefore rewrite our expression such that π₯ minus four is equal to three to the power of four. The right-hand side is equal to 81. We can then add four to both sides of this equation such that π₯ is equal to 85. The solution set of the equation log base three of π₯ squared minus five π₯ plus four is equal to four plus log base three of π₯ minus one is 85. This is the single solution to our equation.