Question

log(x-2)+log(x+1)=2

Answer 1

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

Answer 2

Expand (x-2)(X+1) first.

Answer 3

\[log(x^2-x-2) =2\]

Answer 4

it also says dont forget to check for extraneous solutions

Answer 5

do I use the quadratic?

Answer 6

log_e (x^2 - x - 2 ) = 2 => x^2 - x -2 = e^(2)
then use the quad. formula.

Answer 7

can you show me?

Answer 8

i think that may be log base 10 (common log) ???

Answer 9

what is?

Answer 10

ur original problem... mimi used base e but i think it is base 10

Answer 11

yeah it is log

Answer 12

nevertheless... the procedure is still the same...

Answer 13

Just use the quadratic formula..after that..

Answer 14

yep...^^

Answer 15

dont I have to get one side equal to zero first?

Answer 16

yes... you do....

Answer 17

drag the left hand side to the right

Answer 18

log(x^2-x-4)=0

Answer 19

right?

Answer 20

\[\large \log(x^2-x-2) =2\]

You can't subtract 2 as is... you first have to put both sides to the power of 10.

Answer 21

so 2^10? 1024?

Answer 22

http://www.mathway.com/answer.aspx?p=calg?p=log%28x-2%29+log%28x+1%29SMB012?p=206?p=?p=?p=?p=?p=0?p=?p=0?p=?p=?p=SolveSMB15theSMB15Equation

Answer 23

do you know how to get that answer though?

Answer 24

i just went 2 the website , typed it in, then told what i want 2 do with it and it gave me a straight up answer :P :P

Answer 25

Yes, put both sides to the power of 10, to remove the logs... like this using log rules

\[\LARGE \log _{a} b = x \]
\[\LARGE a^x = b\]

Answer 26

yeah you have to actually know how to do it...

Answer 27

\[\large \log _{10}(x^2-x-2) =2 \]
Try using the rule above.

Answer 28

okay then multiply it out to get x^2-x-2=1024

Answer 29

Not 2^10...
\[\LARGE \log _{a} b = x \]then\[\LARGE a^x = b \]

\[\LARGE \log _{10}(x^2-x-2) =2\]

Answer 30

10 to the second?

Answer 31

Yep.

Answer 32

x^2-x-102=0