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Solve each equation for x. (a) $ \ln (\ln x) =…

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Problem 53 Hard Difficulty

Solve each equation for x.

(a) $ 2^{x - 5} = 3 $
(b) $ \ln x + \ln (x - 1) = 1 $

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Video Transcript

All right here we're gonna have fun solving two equations for X. And look we have all sorts of exponential and log functions. All right, here we go. Starting with a we have two to the X -5 equals three. We want to get that X -5 out of the exponent. We can do that if we do. Um basically if we make a composited inverse functions we can pop out the argument. So the inverse function of two to something is log base two. So we're gonna take log base two of both sides of the equation. So this will give me log base two of three on the right side. Okay that does I wanted that to be a two but it didn't look very much like a two. So let's fix this one down here. There we go. That's log base two. So log base two of two to the X minus five should equal log base two of three. These are inverse composite functions. So out pops my argument. Okay, so I'm almost done. Really, all I have to do is add five, We'll add 5 to both sides. And our answer then for X will be logged Base two of 3 plus five. So that is our solution. Alright? There it is. Okay, next let's look at B. B. Has logs and we've got to terms with X. So what we can do is actually do a log roll when I'm adding two logs of the same base and I can multiply the argument. So this will be equal to Ln of x times x minus one. And we can keep that equal to one to get the excess out of the argument of L. And we do uh the basically we want to create a composite of inverse functions and what we'll do that E to the L. N. Those are inverse. So we're gonna do E to the L N FXX -1. And that will equal E. To the one we do eat at both sides. These will undo each other out. Pops the argument. And the argument, by the way, is expert minus X. If you distribute okay that equals E. So let's go ahead and we have to solve for X. So we actually have to solve a quadratic. Well let's go for it. So what we're gonna do is to attract E from both sides and solve this. This is not one. We can easily factor. So we're going to use the quadratic formula. Here we go. So extend will equal minus P. So minus that. So one plus square root of let's make a little have a little bit more room square root of B squared to minus one squared minus four. A. Is one. C. Is a minus E. And then over to A. So to Alright, so X N is equal to one plus or minus. Well it's not going to think that much, it's really going to just be one plus four E. In there over to. So um uh Normally you would say you have both, the plus and minus are fine but we have some constraints. Um basically L. N. Has constraints on it. For L. N. Um the argument must be greater than zero, so X has to be greater than zero, but this is even more of a constraint. So in this case X has to be greater than one. So we need X to be greater than one, which means we don't want to be subtracting off this value, we need uh we need this to be greater than one. So let's pick then one plus square root of one plus four E. Over to. We have to make sure the square root of one plus four E. Is bigger than one because we do need X to overall Be bigger than one. And it is because four e. You know, E is like 2.7. So square rooting something that's 12 ish. Yeah, definitely. We're good. So this X is greater than one and this then is our solution uh to our equation. So we did it. Congratulations. Hopefully that helped have a wonderful day.

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Solve each equation for x. (a) $ \ln (\ln x) =…

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Michael Jacobsen

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Michael Jacobsen

Joseph Lentino

James StewartCalculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Michael Jacobsen

Joseph Lentino

James Stewart

Calculus: Early Transcendentals