Solve the systems in Exercises 79–80. log x^2=y+3, log x=y−1

Question

Solve the systems in Exercises 79–80. log x^2=y+3, log x=y−1

Accepted Answer

Welcome back. I am so glad you're here. We are given this system of equations. The first equation is log of Y, cute equals x minus two. And the second equation is the log of why Equals X -4. And we are asked to solve for the system of equations. Well there are several ways to go about doing this. I'm going to solve it using substitution. I'm going to add four to both sides of the second equation so that I'll get X by itself. I'm just going to switch sides on these x equals and then it will be log of. Why? That's a four log of y plus four. And now I can substitute where I have an X. In the first equation, I can substitute this log of Y plus four and then solve for why? So now the first equation will read log of why cute equals no longer X. But now log Of Why Plus four. And remembering that -2. Now I'd like to do a couple of things, I would like to bring this log of Y over to the left hand side so that the logs are together. We can see if we can combine them. And then to combine these like terms of plus four minus two. So the left side will read log of y cubed minus the log of why Equals for -2 is two. And now we recall from previous lessons on combining logarithms that when we're subtracting two, logs that have the same base, then we can combine them and we have the log of what's inside the parentheses divided by each other. So the first one goes in the numerator, why cubed divided by the second one? Why That still equals 2? And then we can simplify what's in the parentheses there, that'll be the log of well why cubed divided by? Why is why squared? So we have the log of y squared equals two. And now we can recall some more things about logarithms. We know that the log B of X equals Y and simultaneously be raised to the Y equals X. So that's essentially how we kind of get rid of the log when it's set up like the first one we can switch it to the format of the second equation. And yes, this does look like this first one. But let's identify what we can. What's the B term here will be is unwritten here. But when it's unwritten we know that that is a 10. So B equals 10. R X is what's inside the parentheses. So for this X equals y squared, going back and forth between our Xs and wise because over here on the right hand side of the equal side, that's R Y. So y equals two. And now we can transform it to this equation. So RB is tense will have 10 raised to the Y are y is two, so 10 squared equals and r X is y squared. So now we have 10 squared equals y squared, We still want to get that Y by itself will take the square root of both sides. And we get why equals plus or minus 10 squared of 10 squared is plus or minus 10. Except we also know that we have a domain restriction with logs that anything inside the parentheses for our logs. So in this case why and why cubed? They have to be greater than zero. So we can reject This negative value and we have only y equals the positive 10. So for our solution set we know our Y value that's going to be 10 and we just have to solve for X. We can choose either equation to solve for X. I'm going to use the second one. So log of no longer. Why? But now 10 Equals X -4. And that log of 10. While we can again recall from previous lessons on the properties of logarithms that when we have log B of B when that base is the same as what we're taking the log of that equals one. And as we just said before that unwritten base is 10. So this is log 10 of 10 which equals one. So we can replace that left hand side with 11 equals x minus four, solving for X adding four. Table sides will one plus four equals five. So X equals five. We look at our answer choices and this matches with answer choice. D Well done. We'll catch you on the next one

Solve the systems in Exercises 79–80. log x^2=y+3, log x=y−1

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