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Solve the logarithmic equation for . Please round the answer to four decimal places, if necessary. ln(x-4)+ln(x+5)=1

Mathematics
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Reason I'm confused is because unlike log, in ln the exponential form is e^y=x and I'm having trouble solving the rest
ln(a) + ln(b) = ln(ab)
That's x^2+x-20?

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Other answers:

yep. now, ln(0) = 1 because e^0 = 1
and also any number to the 0 power except 0 is 1
your equation is really: x^2 +x - 20 = 0
wait!
or (x-4) (x+5) = 0
e^1=e
we have (x-4)(x+5)=e
lol..... maybe it does, and maybe it doesnt :)
yes myininaya
if we do it like amistre we would make the problem much easier but also kindof wrong
and kindof wrong is just wrong
now im confused >_<
horseshoe and handgranades me friend :)
so we have x^2-9x+20+e=0
oops -e
wouldnt it be -e?
Yes i blame amistre
lol
bring over 1 to the left at the start...before changinglg form...
...its my liver pills...blame the liver pills :)
so whats the answer?
use quadractic formula x=(9@sqrt(81-4(1)(20-e))/2 where @ means + or -, but make sure your answer is positive because you cannot do ln(negative)
sorry...doesn't make a difference what I said
Quadratic formula the way to go
this is what i put on the formula -1+squareroot(1^2 -4(1)(20-e))))/2
no
a=1 n=-9 c=20-e correct?
c=-20-e
b=-9*
b=1
no b =1
dont listen to me
im doing two things at once sorry
my calculator wont work
(-1+sqrt(1-4(-20-e))/2=4.2925 and (-1-sqrt(1-4(-20-e))/2=-5.2925 but we can't use -5.2925 since it is negative and the natural log of a negative number doesnt exist
x can be negative if it is not the only thing controling the value of the log. the thing is: x-4 needs to be greater than 0 and, x+5 needs to be greater than 0. x >4 serves both cases.
correct!
that negative number gives you another negative number when you plug it into x-4

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