Video: Solving Logarithmic Equations

Find the set of all π‘₯ such that ln (βˆ’8π‘₯ βˆ’ 7) + ln (βˆ’6π‘₯ βˆ’ 1) = ln 7.

04:53

Video Transcript

Find the set of all π‘₯ values such that the natural log of negative eight π‘₯ minus seven plus the natural log of negative six π‘₯ minus one equals the natural log of seven.

We’re given this equation, and the first thing we wanna do is try and see if we can simplify what’s on the left side of the equation. We know that the natural log of π‘₯ plus the natural log of 𝑦 is equal to the natural log of π‘₯ times 𝑦. And so, we can simplify this to say the natural log of negative eight π‘₯ minus seven times negative six π‘₯ minus one. And then, we have one natural log value on either side of the equation. What we wanna do now is take 𝑒 on either side of the equation and make the expression of either side the exponent of the 𝑒 value.

𝑒 to the natural log of negative eight π‘₯ minus seven times negative six π‘₯ minus one is equal to what’s inside the natural log exponent: negative eight π‘₯ minus seven times negative six π‘₯ minus one. In the same way, the other side of the equation will be what’s inside the natural log exponent, which is just seven. At this point, we’re in a position to solve for π‘₯. But before we do that, we need to FOIL or distribute across the parentheses.

Negative eight π‘₯ times negative six π‘₯ equals 48π‘₯ squared. Negative eight π‘₯ times negative one is positive eight π‘₯. Negative seven times negative six π‘₯ equals positive 42π‘₯. And negative seven times negative one equals positive seven. All equal to seven. If we combine like terms, we get 48π‘₯ squared plus 50π‘₯ plus seven equals seven. And then, we want to set this expression equal to zero, so we subtract seven from both sides. And we get 48π‘₯ squared plus 50π‘₯ equals zero. The positive seven minus seven cancels out. Seven minus seven equals zero.

But remember, our goal here is to find the values of π‘₯. I noticed that there is an π‘₯ in either term, and also both of the coefficients are divisible by two. If we take out a factor of two π‘₯, we’ll be left with 24π‘₯ plus 25. And then, we’ll set each factor equal to zero. Two π‘₯ equals zero. 24π‘₯ plus 25 equals zero. On the left, if we divide both sides by two, we get π‘₯ equals zero. And on the right, we subtract 25 from both sides and we get 24π‘₯ equals negative 25. And then, we divide both sides by 24 to get π‘₯ equals negative 25 over 24.

Now, we have two values for π‘₯, but we need to be careful. Remember that our π‘₯s would be plugged back in to a natural log function, and these cannot be negative. We cannot take the natural log of a negative value. So, if we take the expression negative eight π‘₯ minus seven and we plug in zero, that value must not be negative, therefore must be greater than zero. But negative eight times zero minus seven is negative seven. And negative seven is not greater than zero, which means π‘₯ cannot be equal to zero. When π‘₯ is equal to zero, the natural log of negative eight π‘₯ minus seven is undefined. And so, it’s not a valid solution.

We can then check negative 25 over 24. We plug that value in for π‘₯ and we have negative eight times negative 25 over 24 minus seven. This value needs to be greater than zero for it to be a valid solution. When we solve that, we get four-thirds. Four-thirds is greater than zero. And so, we’ll check the second expression; negative six times negative 25 over 24 minus one needs to also be greater than zero. When we solve that value, we get 21 over four, which is greater than zero. And it means π‘₯ equal to negative 25 over 24 is a valid solution. The set of π‘₯ values is only negative 25 over 24.

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