Question

Will Medal

Write the following expressions as a single logarithm...

Answer 1

\[2\log_{b} q+8\log_{b} t\]

Answer 2

\[4\log x -6\log(x+2)\]

Answer 3

Show work pls....

Answer 4

@iGreen

Answer 5

@Nnesha

Answer 6

@confluxepic

Answer 7

Luckily I just learned this.

Answer 8

\[\huge\ y\log_{b}x+z\log_{b}t=\log_{b}x^{y}\log_{b}t^{z}\]

Answer 9

@omarbirjas

Answer 10

Just plug in the values and you'll be fine.

Answer 11

Thanks m8....

Answer 12

You're welcome.

Answer 13

\[\huge\log_{b}x^{y}\log_{b}t^{z}=\log_{b}{x^{y}}{t^{z}}\]

Answer 14

That's the correct version.
If it's a (-) then you divide. If it's a (+) then you multiply.

Answer 15

ok that got me set for the first one....but how do you do the second one...\[\log_{b} q^2t^8\] right?

Answer 16

and for deviding it would look like this? \[\frac{ x^y }{ t^z }\]

Answer 17

@confluxepic

Answer 18

Yes.

Answer 19

Should I show you how to do the second one?

Answer 20

yes please, dont give me the answer tho....

Answer 21

\[\huge\ 4\log x -6\log(x+2)\]
You already know that \(\ 4\log x\) become \(\log x^{4}\).

Answer 22

So you do the same exact thing with \(\ 6\log(x+2)\).

Answer 23

It becomes \(\log(x+2)^{6}\).

Answer 24

\(\huge\ log x^{4}-log(x+2)^{6}\)

Answer 25

There is a minus sign so what do you do?

Answer 26

I posted the rules earlier.

Answer 27

\(\color{blue}{\text{Originally Posted by}}\) @confluxepic
If it's a (-) then you divide. If it's a (+) then you multiply.
\(\color{blue}{\text{End of Quote}}\)

Answer 28

We devide....but its the parethesis that have me goggley eyed...

Answer 29

Oh yeah.

Answer 30

\(\huge\ (x+2)^{6}\)

Answer 31

You can easily square it.

Answer 32

or hex it....lol.....\[(6x+64)\] right? but then what?

Answer 33

Can you explain me the step you took to get that?

Answer 34

\(\huge\ (a + b)^2=a^2+2ab+b^2\)

Answer 35

Replace the square with 6.

Answer 36

\(\huge\ (a + b)^6=?\)

Answer 37

\[64+192 x+240 x^2+160 x^3+60 x^4+12 x^5+x^6\] im guessing?

Answer 38

I got \(\ x^6+12x^5+60x^4+160x^3+240x^2+192^x+64\).

Answer 39

Well they are the same thing.

Answer 40

\(\ log x^{4}-log(x+2)^{6}=log\frac{x^4}{x^6+12x^5+60x^4+160x^3+240x^2+192^x+64}\)

Answer 41

Do you understand?

Answer 42

You said earlier that we had to divide it so I'll take it as a yes.

Answer 43

Yea I got it, now just simplify right?

Answer 44

Yes.

Answer 45

would it be this?? \[24\log_{(\frac{ x }{ x+2 })} \]