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Vampire

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I need help on this q:

i. Differentiate with respect to x the function h(x)=(log_10 x)/x

ii. Given that the only stationary point of h(x) is a maximum, deduce e^pi>pi^e

Thx
 

香港!

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Vampire said:
I need help on this q:

i. Differentiate with respect to x the function h(x)=(log_10 x)/x

ii. Given that the only stationary point of h(x) is a maximum, deduce e^pi>pi^e

Thx
h(x)=(log_10 x)\x=(ln x)\x(ln 10)
h'(x)=(1\ln10) [1-lnx]\x²
Set x'=0
(1\ln10) [1-lnx]\x²=0
1=lnx
.: x=e
At x=e, y=h(x)=1\e(ln10)
This y is "maximum"
This y value would be greater than the y value at x=pi
So at x=pi, y=(ln pi)\pi (ln 10)

1\e(ln10)>(ln pi)\pi(ln10)
1\e>(ln pi)\pi
pi\e>ln pi
e^(pi\e)>pi
put each one to the power of e
e^pi>pi^e as required
 

Slidey

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h(x)=(log_10 x)/x=lnx/(xln10)=y
lny=ln(lnx)-ln(xln10)
y'/y=1/x.1/(lnx)-1/x=(1-lnx)/(xlnx)
y'=(1-lnx)/(xlnx)*lnx/(xln10)
h'(x)=y'=(1/ln10)(1-lnx)/x^2
 

Stefano

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Slide Rule said:
h(x)=(log_10 x)/x=lnx/(xln10)=y
lny=ln(lnx)-ln(xln10)
y'/y=1/x.1/(lnx)-1/x=(1-lnx)/(xlnx)
y'=(1-lnx)/(xlnx)*lnx/(xln10)
h'(x)=y'=(1/ln10)(1-lnx)/x^2
You lost me at y'/y.

Refresh my memory:
If y=ln[g(x)]
then y'= ?
 

dawso

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Vampire said:
I need help on this q:

i. Differentiate with respect to x the function h(x)=(log_10 x)/x

ii. Given that the only stationary point of h(x) is a maximum, deduce e^pi>pi^e

Thx
part (i) - by "steele's method"
 

dawso

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Stefano said:
You lost me at y'/y.

Refresh my memory:
If y=ln[g(x)]
then y'= ?
y' = g'(x) / g(x)
 

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