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

Find the solution set of logβ‚‚ π‘₯ + 9 log_π‘₯ 2 = 6 in ℝ.

03:23

Video Transcript

Find the solution set of log base two of π‘₯ plus nine log base π‘₯ of two equals six in the set of real numbers, 𝑅.

We have an equation. And we’re looking for the real values of π‘₯ which satisfy this equation. The difficulty here is that the two logarithms we have have different bases. While π‘₯ is the argument or input to the first logarithm with base two, it’s the base of the second logarithm.

We’d much prefer it if these two logarithms had the same base. So either we should rewrite this logarithm in terms of the logarithms base π‘₯ or we should rewrite this logarithm in terms of logarithms base two.

On the basis that we’d rather have an equation with log base two in at the end, we’re going to rewrite log base π‘₯ of two in terms of log base two. We’re changing the base of this logarithm. And so we’re going need to change of base formula; that is, that log base 𝑏 of π‘₯ is equal to a log base π‘Ž of π‘₯ over log base π‘Ž of 𝑏 for any value of π‘Ž.

We want to change the base of log base π‘₯ of two. Confusingly our value of π‘₯ is two. And so in the numerator, we have log base π‘Ž of two, while our value of 𝑏 is π‘₯. So in the denominator, we have log base π‘Ž of π‘₯. We’re now are free to pick a value of π‘Ž as this equation is true for all values of π‘Ž.

We choose to set π‘Ž equal to two because we wanted to rewrite this term in terms of log base two to match up with the other logarithm in our equation. And having set π‘Ž equal to two, we can notice that the numerator log base two of two, which is just one. The logarithm to any base of that base is going to be one.

And so we have this very pleasing fact that log base π‘₯ of two is equal to one over log base two of π‘₯. We’ve managed to rewrite log base π‘₯ of two in terms of a log base two. And so now we just need to substitute this expression into our equation.

Doing so, we get this equation. Multiplying both sides of the equation by log base two of π‘₯, we get a quadratic in log base two of π‘₯. Subtracting six log base two of π‘₯ from both sides, we get this quadratic in the standard form we used to.

We can factor this quadratic in the normal way. You might feel more comfortable introducing the variable 𝑦 to represent the log base two of π‘₯ and then factoring. Upon going back to writing things in terms of log base two of π‘₯, instead of 𝑦, you get the equation we have.

In any case the only possible value of 𝑦 or log base two of π‘₯ is three. We raise two to the power of both sides. We use the fact that two to the power of log base two of π‘₯ is π‘₯. In fact this is true for any base π‘Ž, not just two. We find that π‘₯ is equal to two to the three or eight.

As we’re looking for the solution set of the equation and not just the solutions to the equation, we need to put our solution eight into a set to get our final answer. The solution set of log base two of π‘₯ plus nine log base π‘₯ of two equals six in the real numbers is just the set of eight.

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