Question

solve ln3-ln(x+5)-lnx=0

Answer 1

\[ \ln 3-\ln(x+5)-\ln x=0\]try to first combine this into one log like we did earlier, what do you get?

Answer 2

ln 3
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x+5 / x ??

Answer 3

almost, but realize that\[\ln({3\over x+5})-\ln x=\ln({3\over x^2+5x})=0\]you just need to be more careful with the algebra.
After you get that, raise the log's base to the power of both sides. Can you do that?

Answer 4

If this step is giving you trouble, look at just the algebra:\[{{3\over x+5}\over x}={3\over x+5}*{1\over x}={3\over x(x+5)}={3\over x^2+5x}\]

Answer 5

ok seeing it like that is much better, so from there i can just factor out?

Answer 6

form here we have\[\ln({3\over x^2+5x})=0\]do you have any idea how to proceed from here?
(hint: it's not factoring)

Answer 7

quadratic formula?

Answer 8

TuringTest: I admire you, taking your time to explain! Nice going!

Answer 9

Not yet, we need to get rid of the log first. So we raise both sides to the power of the base, because\[a^{\log_a x}=x\]
so, first question: what is the base if this log? (i.e. what is "a" for us?)
@alifetouni thanks :)

Answer 10

Sorry, typo, we raise the base of the log to the power of both sides :/

Terminology can be a little tricky with logs...

So the question I am asking is what is the base of the logarithm \[\ln x\]?

Answer 11

is it e

Answer 12

yes!
now what is\[e^{\ln({3\over x^2+5x})}=e^0\]?

Answer 13

well the ln and e = 1 correct? so then you just have 3/x^2+5x=1 because anything raised to 0 is 1 right?

Answer 14

perfect :)
so now we have\[{3\over x^2+5x}=1\]can you solve this quadratic equation for x?

Answer 15

umm im not to sure haha let me try

Answer 16

Go for it!
Hint: Try to get it looking like\[ax^2+bx+c=0\]then you can either factor it or use the quadratic formula.

Answer 17

so i got x^2+5x-3=0

Answer 18

good!
This is not easily factorable, so can you apply either the quadratic formula or complete the square?

Answer 19

-5+or-√37
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2

Answer 20

That's what I got =)
Good job, hope that helped. I have to go for a bit.
ttyl...

Answer 21

ok thanks a bunch!