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Solve each equation for x.

(a) $ e^{7 - 4x} = 6 $(b) $ \ln (3x - 10) = 2 $

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01:37

Jeffrey Payo

02:11

Heather Zimmers

Calculus 1 / AB

Calculus 2 / BC

Calculus 3

Chapter 1

Functions and Models

Section 5

Inverse Functions and Logarithms

Functions

Integration Techniques

Partial Derivatives

Functions of Several Variables

Johns Hopkins University

Oregon State University

Harvey Mudd College

University of Nottingham

Lectures

04:31

A multivariate function is a function whose value depends on several variables. In contrast, a univariate function is a function whose value depends on only one variable. A multivariate function is also called a multivariate expression, a multivariate polynomial, a multivariate series, or a multivariate function of several variables.

12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

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05:45

All right here we have a couple of equations to solve for X. Let's get started. Alright for question A. We have E. To the seven minus four X equal to six. We want to get that X. Out of the expo net. So we can do that by taking natural log of both sides. So natural log of E to the seven minus four X equals natural log of six. Well natural log of E to the seven minus four X is just seven minus four. X. You can look at that by kind of definition of how logs work but you can also notice that Ln of E. Those are composite inverse functions and out pops the argument. So there we have it basically they undo each other and the argument comes out Alright so now let's go ahead and add four X to both sides. Well actually let's subtract seven instead let's go ahead and subtract seven from both sides. And then we can divide both sides by negative four. So L. N S six minus seven over minus four is what X equals. And if I want to go ahead and multiply top and bottom by a negative one then I can rewrite it as seven minus Ln S. Six. All over four. So there we have it, we solve for X. And part A. So let's go ahead and do part B. In this case I want to get the X. Out of my argument of L. N. So I'm gonna do a similar trick. Remember composite inverse functions out pops the argument. So my trick is that I'm gonna take it to both sides. If I take E to both sides, then again these undo each other out potsie argument three x minus 10 equals E squared I can add 10 to both sides, so I'll get E squared plus 10 and finally divide by three. So X. Is E squared plus 10 all over three. Uh So I think we have successfully solved both equations for X. Alright, hopefully that was fun. Have a great day. See you next time.

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