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

# Question Video: Finding the Solution Set of Logarithmic Equations over the Set of Real Numbers Mathematics • Second Year of Secondary School

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Find the solution set of log_π₯ (5) + log_π₯ (40) β 2 log_π₯ (4) = 2 + log_π₯ (8) in β.

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, weβll copy down our equation. And though there are a few strategies we could use, letβs just start by working from left to right. So we have log base π of π₯ plus log base π of π¦, which will be equal to log base π of π₯ times π¦, 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 weβll just bring everything else down.

Weβ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 π times log base π of π₯ is equal to log base π of π₯ to the π power, which means two times log base π₯ of four is equal to log base π₯ of four squared. And then weβ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 thatβ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 weβ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 π of π₯ equals π, then we can rewrite that as π₯ equals π 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 weβ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 donβ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.

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