Question

Solve: log2(6-2x)-log2x=3

Answer 1

use this property
\(\Large \tt\color{purple}{loga+logb=loga.b}\)

Answer 2

I don't really get it...

Answer 3

well a in your case is 2(6-2x) and b is 2x

Answer 4

Oh im sorry error lol

Answer 5

you need to use this property
\(\Large \tt\color{purple}{log(a)-log(b)=log(\frac{a}{b})}\)
i missed the minus sign xD

Answer 6

so \(\Large \tt\color{purple}{log[2(6-2x)]-log(2x)=3\\log[\frac{2(6-2x)}{2x}]=3\\log[\frac{6-2x}{2x}]=3}\)
carry on the rest

Answer 7

correction!
\(\Large \tt\color{purple}{log[\frac{6-2x}{x}]}\) i forgot to cancel that 2 from top and bottom

Answer 8

@AnnGrey

Answer 9

I'm supposed to solve for x right?

Answer 10

correct! what can you do next?

Answer 11

Multiply? I don't remember my math anymore...I lost it during the summer.

Answer 12

No you have to do inverse operation of log
which exponential

Answer 13

by any chance, is x = 6/(2+e^3)

Answer 14

check and see, do you know how to undo log?

Answer 15

What is all this? Are we wandering in the wilderness?

Are these logs Base 2?

log2(6-2x)-log2x=3

log2((6-2x)/x) = 3

2^3 = (6-2x)/x

8x = 6-2x

10x = 6

x = 6/10 = 3/5

Checking...

log2(6-2(3/5))-log2(3/5)=3

log2(30/5 - 6/5)-log2(3/5)=3

log2(24/5)-log2(3/5)=3

log2((24/5)/(3/5))=3

log2((24/5)*(5/3))=3

log2((24/3)*(5/5))=3

log2((8)*(1))=3

log2(8)=3

Yup!

Answer 16

oh i thought it wasn't the base -_-

Answer 17

No worries. This notation is always difficult to translate.

Answer 18

^_^ thanks though for checking