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Show that $\ln (a x+b)=\ln (x+b / a)+\ln a,$ and therefore $y=\ln (a x+b)$ is a translation of $y=\ln x.$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 6

Properties of Logarithmic Functions

Campbell University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

Lectures

01:23

Show that $\ln (a x+b)=\ln…

01:07

Prove that $\ln x^{y}=y \l…

02:22

verify that $w_{x y}=w_{y …

01:10

Use implicit differentiati…

02:00

Simplify:$$\ln a-\ln b…

01:36

Show that the inverse of $…

Okay here, I want to show that this statement is true. So let's begin off then with the left hand side at hs I can write that as Ln of A, times X plus be over right, same thing. And then by the first law of logs that can write that as Ln of A plus Ln of X plus B over A. And that is it basically just rewrite it Ln of X plus a few of array Plus Ln of eight that becomes therefore a translation and the X movement is negative B over A and the Y movement is up and then A. So the graph of Ln X is translated by this factor, which means it's going left B over A and up. Hello Nate.

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