Question

please help solve:
log(x)+log(x-14)=log(20x)

Answer 1

start with
\[\log(x(x-14))=\log(20x)\] then
\[x(x-14)-20x\] solve the quadratic

Answer 2

no times, not plus

Answer 3

\[\log(x)+\log(x-14)-\log(20x)=0\]
\[\log(\frac{x(x-14)}{20x})=\log(1)\]
=>
\[\frac{x(x-14)}{20x}=1 =>x^2-14x-20x=0=>x^2-34x=0=>x(x-34)=0\]

Answer 4

\[x^2+14x=20x\]
\[x^2-6x=0\]
\[x(x-6)=0\]
\[x=6\] is the only solution because you cannot take the log of 0

Answer 5

@myininaya too much work!!

Answer 6

x=6?

Answer 7

i don't think that will work satellite

Answer 8

no i put + when it should have been -

Answer 9

oh you change the problem

Answer 10

anyways x=34 is right

Answer 11

should have been
\[x^2-14x=20x\]\[x^2-34x=0\]
\[x(x-34)=0\]
\[x=34\]

Answer 12

since we have x(x-34)=0
x=0 or x=34
and x=0 is not a solution
x=34 is only solution

Answer 13

but still no need for quotients or the log of 1

Answer 14

satellite likes to make up stuff when he does problems

Answer 15

just thought i would mention it

Answer 16

so its not wrong to do lol

Answer 17

because everyone is a critic, especially me

Answer 18

thank you both of you!

Answer 19

i think we work better together

Answer 20

wonderful conversation, more dramatic than the problem itself

Answer 21

satellite and i like it that way