Question

2lg5-lg(x+2)=1-lg(2x-1)

Answer 1

\[\Large 2\log5-\log(x+2)\quad=\quad 1-\log(2x-1)\]

So we need to remember a couple of important log rules:\[\Large\color{#DD4747 }{\log a-\log b=\log \frac{a}{b}}\]

Answer 2

\[\Large\color{#3399AA }{b \log (a)=\log(a^b)}\]

Answer 3

Understand how we can apply the blue rule to the first term, 2log5?

Answer 4

Is it the change of base law?

Answer 5

no.. :o
it allows us to bring coefficients in front of the log inside as exponents.

Answer 6

Oh wait,it's the quotient law right?

Answer 7

@CalvinGordon the base is always 10

Answer 8

no not the quotient law yet..
can you see the blue one i posted....? :\

Answer 9

Power law?

Answer 10

Yes, that's probably what it's called.
Do you see how we can apply that to change 2log5?

Answer 11

Yes

Answer 12

So it brings the 2 inside as an exponent, right? :o

Answer 13

So it should be \[\log_{}5 ^{2}\]

Answer 14

Ok good,\[\Large \log25-\log(x+2)\quad=\quad 1-\log(2x-1)\]

Answer 15

Looks like we can apply the quotient rule on the left side now, yes?

Answer 16

Yup

Answer 17

Something like this? :o\[\Large \log\left(\frac{25}{x+2}\right)\quad=\quad 1-\log(2x-1)\]

Answer 18

The right side is a little tricky.
You need to recall that if your base ever matches the contents of the log, the result is 1.
So there is a sneaky way we can rewrite that 1.\[\Large \log10=1\]

Answer 19

\[\Large \log\left(\frac{25}{x+2}\right)\quad=\quad \log10-\log(2x-1)\]

Answer 20

@zepdrix you can throw everything except 1 to the left first :p

Answer 21

Wait,is lg and log the same thing?

Answer 22

Ya you could do that also :3 maybe simpler that way.

Answer 23

What is lg...? I assumed it was a typo.. ive never seen lg before.

Answer 24

Except on cellphones.

Answer 25

Well,it's all over my textbook,even the question.It's known as the common logarithm.

Answer 26

Hmm I had to Google it.
I guess the `common logarithm` is the same as log with no subscript, base 10.
So yah we're ok.

Answer 27

Unless it's meant to be log base 2.
Have you dealt with that at all?

Answer 28

Nope,could you give me an example?

Answer 29

@CalvinGordon please give @zepdrix a medal by clicking on the "best answer" button :)
\[2\log5-\log(x+2)=1-\log(2x-1)\\\log25-\log(x+2)=\log10-\log(2x-1)\\\log\frac{25}{x+2}=\log\frac{10}{2x-1}\\\frac{25}{x+2}=\frac{10}{2x-1}\\50x-25=10x+20\\40x=45\\x=1.125\]

Answer 30

This kind of question is just asking you whether you know the log identities and how to use them

Answer 31

Alright,noted.