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Solve for $x.$$$\log x+\log (x+1)=\log 12$$

$$3$$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 6

Properties of Logarithmic Functions

Missouri State University

Campbell University

University of Michigan - Ann Arbor

Idaho State University

Lectures

02:13

Solve for $x.$$$\log (…

01:06

Solve exactly.$$\l…

00:23

Solve$$\log _{x} 1…

01:25

$$\text { Solve } \log (x+…

01:27

Solve each equation for $x…

00:30

02:16

01:05

Solve for $x$ algebraicall…

00:33

Solve the logarithmic equa…

01:04

Solve exactly.$$\log (…

00:24

00:36

01:12

Solve each equation for th…

So we're looking at this problem of log of X for anybody that's confused. It's an implied base of 10. If they don't write it, um, log of X plus wine. Uh, same implied base log of 12 with the same implied base. Eso the main promise of those problems. Understand that when you're adding logs together, then we combine them using multiplication. So I'm looking at log. The base stays the same of X Times X plus one. Now, I would actually distribute that in. That's just me. Um, I'll do that in a second step. Not do anything with the right side. So technically, what I'm doing, Thio, get rid of the logs. I'm taking 10 to that power, so tend to the log based 10. Power will cancel out and saying with this, um, so what I'm left with is that X squared plus X as equal to 12. Well, now it's a quadratic where we can solve quadratic. So we just got to get the equation equal to zero. Subtract 12 over. So now I can factor well factors of 4, 12 or four and negative three. Um, and now I can use a zero product property to say that X is either equal to negative four or because that's the only way to get this piece equal to zero or X equals positive. Three. Now here's the deal. Um, is if I go back and I plugged in Negative four in for log. We can't log in negative four. We can't log any negatives. Eso This answer is extraneous. It does not work. The only answer that works is X equals three and we're done.

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