Video: Finding the Solution Set of Logarithmic Equations over the Set of Real Numbers

Find the solution set of log_π‘₯ (5) + log_π‘₯ (40) βˆ’ 2 log_π‘₯ (4) = 2 + log_π‘₯ (8) in ℝ.

02:55

Video Transcript

Find the solution set of log base π‘₯ of five plus log base π‘₯ of 40 minus two times log base π‘₯ of four is equal to two plus log base π‘₯ of eight in all real numbers.

First, we’ll copy down our equation. And though there are a few strategies we could use, let’s just start by working from left to right. So we have log base 𝑏 of π‘₯ plus log base 𝑏 of 𝑦, which will be equal to log base 𝑏 of π‘₯ times 𝑦, which means we can combine our first two terms and call them log base π‘₯ of five times 40, log base π‘₯ of 200. And then we’ll just bring everything else down.

We’re close to being able to combine log base π‘₯ of 200 and two times log base π‘₯ of four. And we can use the rule that 𝑝 times log base 𝑏 of π‘₯ is equal to log base 𝑏 of π‘₯ to the 𝑝 power, which means two times log base π‘₯ of four is equal to log base π‘₯ of four squared. And then we’re subtracting two logs that have the same base. And we know when that is the case, we can divide each of their values. log base π‘₯ of 200 minus log base π‘₯ of four squared will be equal to log base π‘₯ of 200 over 16.

It seems like that’s all we can do here on the left side of the equation. What we can do now is we can move this log base π‘₯ of eight to the left side of the equation by subtracting log base π‘₯ of eight from both sides. We now have log base π‘₯ of 200 over 16 minus log base π‘₯ of eight. And we’ll use this subtraction rule again, which will be log base π‘₯ of 200 over 16 divided by eight. We know that divided by eight is the same thing as multiplied by one-eighth. And if we do that simplification, we end up with log base π‘₯ of 25 over 16 is equal to two.

And now we want to take this out of logarithm form and put it into exponent form. If we have log base 𝑏 of π‘₯ equals π‘˜, then we can rewrite that as π‘₯ equals 𝑏 to the π‘˜ power. So we have 25 over 16 is equal to π‘₯ squared. By taking the square root of both sides of the equation, we get that π‘₯ is equal to plus or minus five over four, since the square root of 25 is five and the square root of 16 is four. Since we’re only looking for values that are in the set of all reals, we want to say that π‘₯ cannot be negative five-fourths. We don’t wanna deal with a negative base here, as this would yield an answer that is imaginary. And so π‘₯ is only equal to five-fourths, which makes the solution set just five-fourths.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.