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OpenStudy (anonymous):
is it like ln(x-5x)=0 or .....
OpenStudy (anonymous):
hello tell me to understand
OpenStudy (anonymous):
Go to wolframalpha.com and ask the question.
OpenStudy (anonymous):
Mom! There's evil in my closet! EVIIIILLL!!!
OpenStudy (anonymous):
any way x=\[e^{5x}\]
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OpenStudy (anonymous):
you have one variable but in two side with exponential so wolframalpha.com
canot solve it hhhhhhh
OpenStudy (anonymous):
the solution is no solution
OpenStudy (anonymous):
no...
OpenStudy (anonymous):
you cant say x(1-5) cause the x with ln and -5x is alone so you can't say like
ln x-5x=0
ln x(1-5) = 0
ln -4x= 0
-4x = e^0
-4x = 1
x = -1/4
OpenStudy (anonymous):
lnx = 5x
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OpenStudy (anonymous):
so x = e^(5x)
it don't have any solution
OpenStudy (anonymous):
when we say x = e^(5x)
so we need x= number that make the two sides be equal
there is no any number to this
OpenStudy (anonymous):
until if we derive it be 1=5e^(5x)is critical point
so .........
OpenStudy (tkhunny):
That is not the right equation. Please think about the Domain.
\(f(x) = \ln(x) - 5x\) Domain \(x > 0\)
\(g(x) = e^{5x} - x\) Domain \(x \in \mathbb{R}\)
Neither has a solution, f(x) = 0 or g(x) = 0.
OpenStudy (anonymous):
but how much ln (0) =
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