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OpenStudy (ash2326):
@Dodo1 What do you need to do here?
OpenStudy (anonymous):
Logarithmic Derivatives
OpenStudy (anonymous):
f(x)'=
OpenStudy (anonymous):
@Luis_Rivera
jimthompson5910 (jim_thompson5910):
y = 4^(x*log(7,x))
ln(y) = ln( 4^(x*log(7,x)) )
ln(y) = (x*log(7,x))* ln( 4 )
does that help?
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OpenStudy (anonymous):
log 7 *x?
jimthompson5910 (jim_thompson5910):
what do you mean?
OpenStudy (anonymous):
about part of log (7*x)
OpenStudy (anonymous):
is it times?
jimthompson5910 (jim_thompson5910):
oh log(7,x) means "log base 7 of x"
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jimthompson5910 (jim_thompson5910):
just shorthand notation
OpenStudy (anonymous):
Thats cool, can i ask you other question?
jimthompson5910 (jim_thompson5910):
sure
OpenStudy (anonymous):
One more thing,
so this is homework called webwork online Homework,
I entered the answer it says log is too many inputs
what does it mean
jimthompson5910 (jim_thompson5910):
show me with a screenshot please
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OpenStudy (anonymous):
screenshot?
jimthompson5910 (jim_thompson5910):
take a picture of the screen using the printscreen key so I can see the error message
OpenStudy (anonymous):
sure,
jimthompson5910 (jim_thompson5910):
ok so the log isn't in the exponent at all
jimthompson5910 (jim_thompson5910):
you would use the product rule in this case
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jimthompson5910 (jim_thompson5910):
y = 4^x*log(7,x)
y' = 4^x*ln(4)*log(7,x) + 4^x*1/x*1/ln(7)
y ' = (4^x*ln(4)*ln(x))/ln(7) + (4^x)/(x*ln(7))
jimthompson5910 (jim_thompson5910):
you can optionally factor out the GCF to get
y ' = ( (4^x)/ln(7) ) * (ln(4)*ln(x) + 1/x )
OpenStudy (anonymous):
YES! you are RIGHT!.
I have been asking this for about 4 hrs and people helped me but it didnt work. THANK YOU.
can i ask 5 more questions ? its homework due tomorrow.
OpenStudy (anonymous):
i will open up the new post so you can have medal?
jimthompson5910 (jim_thompson5910):
the system didn't like the log(7,x) since it probably wanted ln(x)/ln(7) or log(x)/log(7)
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jimthompson5910 (jim_thompson5910):
but I think log(7,x) is much more compact
OpenStudy (anonymous):
Yes, Thank you so much!
Do you have time for other questions?
jimthompson5910 (jim_thompson5910):
just ask them here, it's fine
jimthompson5910 (jim_thompson5910):
yeah if they're not too long
jimthompson5910 (jim_thompson5910):
5 shouldn't be too bad
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