Question

can someone help me with this question log(x+3)+log (x)=1

Answer 1

sorry

Answer 2

base 10?

Answer 3

yes

Answer 4

log(x+3)+logx=1

Answer 5

log[(x+3)*(x)]=1

Answer 6

oh yeah. rewrite as one log
\[log(x(x+3)) = 1\] so \[log(x^2+3x)=1\] then rewrite in equivalent experiential form
\[x^2+3x=10^1\]

Answer 7

now you have a quadratic to solve
\[x^2+3x-10=0\]
\[(x+5)(x-2)=0\]
\[x+5=0\]
\[x=-5\] or \[x-2=0\] \[x=2\]

Answer 8

5 and 2 i think are better options ;)

Answer 9

but you cannot take the log of a negative number so answer is 2

Answer 10

no i like my answers. it does factor as
\[(x+5)(x-2)\] right?

Answer 11

yeah

Answer 12

ok thanks i got that

Answer 13

whew so answer is 2

Answer 14

don't put -5!

Answer 15

yeah

Answer 16

i didnt

Answer 17

5 and 2 are better options for the factoring than 5 and 3 :)

Answer 18

can u help me with this one to

Answer 19

did i put 5 and 3?

Answer 20

....no, the deleted post did tho

Answer 21

no i did

Answer 22

oh...

Answer 23

you are paying too close attention!

Answer 24

lol

Answer 25

just trying to keep my mind of the money pit im driving in

Answer 26

so it is 2

Answer 27

the answer is 2

Answer 28

is it

Answer 29

yes 2

Answer 30

and not -5

Answer 31

because \[log(5)+log(2)=log(10)=1\]

Answer 32

you cannot take the log of a negative number

Answer 33

ok

Answer 34

if i write \[log(-5)=x\] that means
\[10^x=-5\] but
\[10^x\] is always positive so it cannot be -5

Answer 35

ok thanks

Answer 36

welcome