Solve the systems in Exercises 79–80. log_y x=3, log_y (4x)=5

Question

Accepted Answer

Welcome back. I am so glad you're here. We are given these two equations in our system of equations. The first one is log X of y equals five and the second one is log X of nine, Y equals seven. Now to solve for X and Y. We need to recall from previous lessons on logarithms how we can rewrite this equation without using log. So we know that when we have the log B of X, that is equal to Y and at the same time be raised to the Y equals X. So that second equation in blue, that's how we write it without the log. So we have to identify R. B X and Y parts for each equation and then rewrite it as B raised to the Y equals X. I'm going to go through this step by step. It might be a little bit confusing because we're going to be changing bees into wise and X is all over the place so I'll write it out and it's okay if you need to pause and really focus on what I'm doing. So I'm going to identify the B, the X and the Y. For equation one. So b corresponds with the base. So the base here is X. X corresponds to what's in the parentheses X. Here is Y. And why corresponds to what it's equal to by itself on the right hand side of the equation. So that is five. It's not always on the right hand side of the equation, but on this case it is so we've identified our B, X and Y and now we can rewrite equation one. Like the second equation in blue where we've got B. So B is X Raised to the Y. Y is five equals X and X is Y. Alright we got one of them. Now let's do the second equation rewriting it will first identify R. B. X and Y terms from this first format. So B. Is the base and in this case B equals X. X. Is what's inside the parentheses here and that equals nine. Y. And why is what's by itself on the opposite side of the equal sign? And in this case y equals seven. And now we're going to rewrite that equation so that it doesn't have the log in it. So it starts with the B. B. Is X Raised to the why? Why is seven equals X. X. is 9? Y. And now this is a bit more doable to solve. We have got this, Y equals in the first equation. And we can use substitution where we have a Y. In the second equation we can substitute in X to the fifth. So rewriting the second equation, that's X to the seventh equals nine times no longer Y. But now X to the fifth from the first equation. And to solve for X let's divide both sides by X to the fifth, get the X's together and then we've got X to the seventh divided by X to the fifth. Well that's just going to leave us with X squared. So we have X squared equals and on the right hand side, nine to solve for X will take the square root of both sides and this will give us X equals plus or minus the square root of nine while the squared of nine is three. And now we take a look back at our original equations and we say that the X. Is the base and we do have a domain restriction there. You can't have a negative base so we can disregard this negative three as an answer and we just have X equals three. Okay, we've got our X value for our solution set and now we just have to solve for Y. We can choose either equation. I'm going to choose equation one. Well, plug in Our x value. So now it's not log base X. It's log base three of why Equals five. And now again we want to get rid of the log and we'll rewrite it so that we don't have the log in there and we're going to rewrite it the same way that we rewrote this first equation here. So instead of X to the fifth that is three to the fifth equals why And 3 to the 5th. That is 243. So y equals 243. We look at our answer choices and this matches with answer choice. C. Well done. We'll catch you on the next one

Solve the systems in Exercises 79–80. log_y x=3, log_y (4x)=5

Key Concepts

Logarithmic Functions

Change of Base Formula

Properties of Logarithms

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