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

Solve log₃(logβ‚„(π‘₯Β² + 39)) = 1, where π‘₯ ∈ ℝ.

02:40

Video Transcript

Solve the equation log base three of log base four of π‘₯ squared plus 39 equals one, where π‘₯ is an element of the set of all real numbers.

We’ve been given a logarithmic equation for which we need to solve for π‘₯. We’ll need to be really careful here. We have two different bases. We’ve got base three and base four for our logarithms. So what is a logarithm? Well, it’s simply another mathematical operation. This time, it determines how many times a certain number ⁠— that’s the base ⁠— is multiplied by itself to reach another number.

Let’s take the general equation log base 𝑏 of π‘Ž equals 𝑐. We raise both sides as a power of 𝑏. And this is really useful because logarithms are the inverse operation to exponentiation. So 𝑏 to the power of log base 𝑏 of π‘Ž is simply π‘Ž. And on the right-hand side, we get 𝑏 to the power of 𝑐. So these two equations are equivalent. Log base 𝑏 of π‘Ž equals 𝑐 is the same as saying π‘Ž is equal to 𝑏 to the power of 𝑐. So let’s go back to our equation. We’ve got log base three of some expression. We’re going to raise both sides of our equation then as a power of three.

When we do, the left-hand side becomes log base four of π‘₯ squared plus 39. That’s the equivalent of π‘Ž in our general form. The right-hand side becomes three to the power of one. That’s the equivalent of 𝑏 to the power of 𝑐. But, of course, three to the power of one is simply three. So our equation is log base four of π‘₯ squared plus 39 equals three. Notice we have another logarithmic equation. This time, it’s base four, so we’re now going to raise both sides as a power of four. Four to the power of log base four of π‘₯ squared plus 39 is just π‘₯ squared plus 39. And our right-hand side becomes four cubed. Of course, we know four cubed is equal to 64, so we have π‘₯ squared plus 39 equals 64.

We’re now going to solve for π‘₯ . Let’s subtract 39 from both sides of our equation such that π‘₯ squared is equal to 25. And then, we’re going to take the square root of both sides, remembering, of course, we need to take both the positive and negative square root of 25. And when we do, we get π‘₯ is equal to plus or minus five. And these are real numbers. So there are two solutions to our equation where π‘₯ is an element of the set of real numbers. It’s five or negative five.

Now, remember, we could also check this solution by substituting π‘₯ equals five and π‘₯ equals negative five into our original equation. In both cases, we do indeed get one, so we know our solutions are correct.

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