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monroe17Best ResponseYou've already chosen the best response.1
\[y=ln(\frac{ x^4e^{3x} }{ \sqrt{2x^51} })\]
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
i think it would be better to do logarithmic differentiation...
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
\(\large y=ln(\frac{x^4e^{3x}}{(2x5)^{\frac{1}{2}}}) = 4lnx + 3x  \frac{1}{2}(2x^51)\) simplifying y that way should be much simpler...
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
i guess that's not so hard afterall....
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
wait wait.. how'd you get 4ln(x)+3x1/2(2x^51)
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
use the logaritm properties...
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
I need to look those up ;/ and review them.
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
\(\large ln(AB)=lnA+lnB \); \(\large ln(A/B)=lnAlnB \); \(\large ln(A^B)=BlnA \); hmm... i think that's all of 'em...
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
Awe, well thank you. Okay so now back to what you got.. Let me look over that real quick
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
yeah... should be easier now....
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
Okay so I see you used the second one.. But, I'm confused how you got those values... ln(A)=4lnx+3x when A= x^4e^{3x}
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
\(\large y=ln(\frac{x^4e^{3x}}{(2x5)^{\frac{1}{2}}}) =ln(x^4e^{3x})ln(2x5)^{1/2} \)
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
that's the first step...
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
now.... \(\large ln(x^4e^{3x})=lnx^4+lne^{3x}=4lnx+3x \) do u undestand this part?
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
ooo... sorry my bad... the second part is wrong....
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
ohhh... ln(A^b)=b*ln(A) ln(x^4)=4*ln(x) and e&ln cancel out leaving 3x
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
\(\large ln(2x5)^{1/2}=\frac{1}{2}ln(2x5) \) that's what it should have been...
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
and yes (to your last post)
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
lemme rewrite what i shouldve written....
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
\(\large y=ln(\frac{x^4e^{3x}}{(2x5)^{\frac{1}{2}}}) = 4lnx + 3x  \frac{1}{2}\color {red}{ln}(2x^51) \)
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
\[\frac{ 4 }{ x }+3..\]
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
I need to do chain rule?
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
yes, chain rule: \(\large [lny]'=\frac{y'}{y} \)
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
so \(\large [ln(2x^51)]'=\frac{[2x^51]'}{2x^51} \) simplify....
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
\[\frac{ 4 }{ x }+3\frac{ 1 }{ 2 }\frac{ 10x^4 }{ 2x^51 } = \frac{ 4 }{ x }+3\frac{ 5x^4 }{ 2x^51 }\]
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
why haha? did I do something wrong?
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
no... i was getting the wrong answers until at the last moment when u posted ur answer... it confirmed my last answer...
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
oh lol :) you just discouraged me! >;[ no im jk ;) thank you!
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
haha... just kidding.... yw... :)
 one year ago

dpaIncBest ResponseYou've already chosen the best response.2
i forgot... nice work on your part... way to stick with the problem....
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
what you mean? I always finish ;D
 one year ago
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