Video: Solving Logarithmic Equations Involving Laws of Logarithms and Quadratic Equations

Solve logβ‚„ ((π‘₯Β² βˆ’ 26π‘₯)/(βˆ’7π‘₯ + 6)) = 1, where π‘₯ ∈ ℝ.

02:25

Video Transcript

Solve log to the base four of π‘₯ squared minus 26π‘₯ over negative seven π‘₯ plus six equals one, where π‘₯ is in the set of real numbers.

Well, the first thing we want to do to solve this problem is use a relationship we know. And that relationship is that log to the base π‘Ž of π‘Ž is equal to one. So, therefore, if we use this, what we can do is rewrite our equation as log to the base four of π‘₯ squared minus 26π‘₯ over negative seven π‘₯ plus six is equal to log to the base four of four. But why would we want to do this? Well, we want to do this because this now has on both sides of our equation log to the same base. So therefore, what we can do is equate the arguments.

So, therefore, we can say that π‘₯ squared minus 26π‘₯ over negative seven π‘₯ plus six is equal to four. So now if we multiply through by negative seven π‘₯ plus six, we’re gonna get π‘₯ squared minus 26π‘₯ equals four multiplied by negative seven π‘₯ plus six. And then if we distribute across our parentheses, we’re gonna get π‘₯ squared minus 26π‘₯ equals negative 28π‘₯ plus 24. So, then, what we can do is add 28π‘₯ and subtract 24 from each side of the equation. And this gives us π‘₯ squared plus two π‘₯ minus 24π‘₯ equals zero.

So, then, to find our values of π‘₯, what we’re gonna do is solve our quadratic. And we can do that using factoring. So then to factor our quadratic, what we’ll get is π‘₯ plus six multiplied by π‘₯ minus four. And that’s because to factor, what we need to do is find two numbers whose product is negative 24 and whose sum is positive two. So, therefore, π‘₯ can be negative six or four because these are the values that satisfy this equation because what we do is look inside our parentheses. And we look for the π‘₯-values that will make each of our parentheses equal to zero.

So, negative six add six would be zero. So, therefore, we’d have a zero multiplied by, and then we’d have π‘₯ minus four. So, zero multiplied by anything is zero. So, it gives us the same answer as we have on the right-hand side of our equation.

So, therefore, we can say the solutions to log to the base four of π‘₯ squared minus 26π‘₯ over negative seven π‘₯ plus six equals one are negative six and four.

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