Question Video: Finding the Solution Set of a Logarithmic Equation over the Set of Real Numbers Mathematics

Find the solution set of logβ‚ˆ √(9π‘₯ βˆ’ 26) + logβ‚ˆ √(π‘₯ +1) = logβ‚ˆ (128) βˆ’ 2 in ℝ.

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Video Transcript

Find the solution set of log base eight of the square root of nine π‘₯ minus 26 plus log base eight of the square root of π‘₯ plus one is equal to log base eight of 128 minus two in all real numbers.

If we copy down our equation, we see that we’re dealing with three terms that are log base eight. But one of our terms is a constant. So let’s focus on the right side of the equation, the side with the constant first. We have a few options, but it’s going to require us to think creatively.

One way of solving this is writing two as a log base eight. We do that by remembering log base 𝑏 of 𝑏 is one. And that means we could rewrite the constant two as two times log base eight of eight. That would be two times one. The other thing we remember is that if we have a constant multiplied by a log base 𝑏 of π‘₯, that’s the same thing as log base 𝑏 of π‘₯ taken to the 𝑝 power. This means the constant two can be rewritten as log base eight of eight squared, which in fact will work very well for us. So we’ll just bring down this log base eight of 128.

Now, if we have log base 𝑏 of π‘₯ minus log base 𝑏 of 𝑦, we can rewrite that as log base 𝑏 of π‘₯ divided by 𝑦. And this means we can rewrite log base eight of 128 minus log base eight of eight squared as a log base eight of 128 divided by eight squared. Eight squared is 64. 128 divided by 64 is two, which means we’ve been able to simplify the log base eight of 128 minus two down to log base eight of two.

From there, we can turn our attention to the left side of the equation. Both of our terms are log base eight. But we’re taking the log of the square root of both of these. But this will be much more helpful if we rewrite the square root as to the one-half power. Once we do that, we’re able to pull out that one-half power so that we now have one-half times the log base eight of nine π‘₯ minus 26 plus one-half times log base eight of π‘₯ plus one. Since one-half is a factor of both of these terms, we can undistribute it.

Now, inside these brackets, we have two logs with the same base that are being added together. And that means we can use the rule log base 𝑏 of π‘₯ plus log base 𝑏 of 𝑦 is equal to log base 𝑏 of π‘₯ times 𝑦. We’ll now have a log base eight of nine π‘₯ minus 26 times π‘₯ plus one. We’ll need to distribute and expand here so that we have nine π‘₯ squared plus nine π‘₯ minus 26π‘₯ minus 26. Nine π‘₯ minus 26π‘₯ equals negative 17π‘₯.

At this point, we’re very close to something we can work with. But we don’t have exactly something we can work with yet because of this one-half. And so what we want to do is actually put this one-half back in exponent form. We want to instead call this the log base eight of nine π‘₯ squared minus 17π‘₯ minus 26 to the one-half power. And hopefully, you’ll see why in just a second. Because if we do that, we now have a log base 𝑏 of π‘₯ equal to a log base 𝑏 of 𝑦, which tells us π‘₯ is equal to 𝑦, which means nine π‘₯ squared minus 17π‘₯ minus 26 to the one-half power is equal to two.

We can get rid of that one-half power by squaring both sides of our equation, which gives us nine π‘₯ squared minus 17π‘₯ minus 26 equals four. If we subtract four from both sides, we get the quadratic nine π‘₯ squared minus 17π‘₯ minus 30 equals zero. And this is something that we can factor to solve for π‘₯.

At this point, we’re not really dealing with any log rules. We’re simply factoring to solve a quadratic. So I wanna find the terms that multiply together. Positive 10 and negative three multiply together to equal negative 30. And negative 27 plus 10 equals negative 17. Setting both of our factors equal to zero, we see that π‘₯ will either be negative ten-ninths or π‘₯ will be equal to three.

However, because of our properties of logs, we know that we can’t take the log of a negative value. If we try to plug in negative ten-ninths into nine π‘₯ minus 26, we would end up with a negative value. And that tells us that negative ten-ninths is not a valid solution for π‘₯. If we use the same method to check for three, nine times three minus 26 is positive. And if we substitute three in for π‘₯ plus one, again, we get a positive value. The solution set for π‘₯ here is just positive three.

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