In Exercises 79–84, solve for y.
83. ln(y-1) = x + ln(y)...

Question

In Exercises 79–84, solve for y.

83. ln(y-1) = x + ln(y)

Accepted Answer

Welcome back, everyone. So for Y in terms of X. LN of Y plus 2 equals X plus LN of Y. So for this problem, let's notice that in this equation we have two natural logarithms. What we can do is rewrite natural logarithms on the same side of the equation. So let's rewrite this equation as LN of Y + 2, subtracting LN of Y from both sides, we get minus LN of Y. Equals x. And now what we can do is apply the properties of logarithms. We have a difference of two natural logs, so the left hand side can be rewritten with a land of the ratio between Y + 2 and a Y, right? Applying the quotient rule of logarithms. This is equal to X. Now, what we can do is take the anti log of both sides. We can express Y + 2 divided by Y by taking E to the power of X. Well done, and now our goal is to solve for Y. So what we can do is apply the properties of fractions on the left hand side, we get 1 plus 2 divided by Y equals e to the power of X. We're going to subtract one from both sides so that 2 divided by Y equals e to the power of X minus Y. and now to solve for Y. We have to divide the numerator by the expression on the right hand side. So Y equals 2, divided by the power of X minus 1. So this is the expression that we want to get and we are going to write it's. Domain. Right? Specifically, since we Have an expression of Y in terms of X, and the original domain must be Y is greater than 0 because we cannot take a n of a number that is less than or equal to 0, so we're taking into account one of the logs, and for the second log Y + 2 must be greater than 0 as well. So solving these inequalities, Y must be greater than 0 and Y must be greater than -2, meaning overall Y must be greater than 0. So to ensure that Y is greater than 0, since our numerator is positive, our denominator must be positive as well, right? And our numerator, I'm, I'm sorry, our denominator is E to the power of X minus 1. It must be greater than 0, which essentially means that it's the power of X must be greater than 1. And this essentially means that X must be greater than 0, because E to the power of 0 is 1, right? So we want to ensure that X is greater than 0. And that's our final answer for this problem. Let's label it and thank you for watching.

In Exercises 79–84, solve for y.83. ln(y-1) = x + ln(y)

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In Exercises 79–84, solve for y.
83. ln(y-1) = x + ln(y)