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Solve for $x.$$$\log (x+1)+\log (x-1)=1$$

$$\sqrt{11}$$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 6

Properties of Logarithmic Functions

Campbell University

Oregon State University

University of Michigan - Ann Arbor

Lectures

01:06

Solve exactly.$$\l…

02:16

01:05

Solve for $x$ algebraicall…

01:58

Solve for $x.$$$\log x…

00:30

01:25

$$\text { Solve } \log (x+…

00:24

Solve the logarithmic equa…

00:33

01:04

Solve exactly.$$\log (…

00:36

02:29

02:24

Solve exactly.$$1-…

00:54

Solve Problems exactly.

All right. So this problem is about inverse functions. Uh, it probably helps to ride out that the implied bases log based 10 of X plus one and then plus log based 10 of X minus one is equal to one eso. The underlying concept that we need to examine is this. Plus, we can combine the logs together as a single log rhythm and, uh, multiplying these together. I hope everybody's familiar that this is a difference of squares that you'll get X squared, the outside term being minus one x. The inside would be plus one x so they'll cancel. And then one times negative one is negative. One is equal to one. I haven't done anything with the right side. So then the next thing is the inverse operation of Log based 10 is taking 10 to that power. That way on the left side, they'll cancel and you're left with X squared minus one. And on the right side tend to the one powers equal to 10, and we're actually almost ready. Toe answer. You know, finalize this question because you just add one over and then when you square root, you do get to answers it's plus or minus the square root of 11. However, if you were to plug in, um, I guess negative Route 11 into the problem right here and subtract one from it. I think you would still get negative over here. Anyway, The reason why negative Route 11 doesn't work is because when you subtract one from a negative, you're definitely going to get a negative, and we cannot log a negative. So if anybody's confused, why we disregard the negative answer in this problem is because we can't log negatives s Let me circle. You're correct. Answer and make that note there you.

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