Intermediate Value Theroem help:
Show that there exista ln^x = e^(-x) on the interval (1,2)

Question

amistre64 · Answer

the notation is confusing

amistre64 · Answer

if you plug in 1 and 2 into the "function" and it produces different signs; then it had to cross at some point in the interval

amistre64 · Answer

as long as the function is continuous

anonymous · Answer

we were supposed to solve it w/out a calulator. somehow we have to know that 1/ln2 is less that e^2 so ln 2 is greater tahn 1

amistre64 · Answer

ln^x = e^(-x) ; this notation makes no sense, specifically that ln^x ... whats it mean?

anonymous · Answer

really...shud it be lnx?

amistre64 · Answer

.... i dunno, i aint got your material to compare with :)

anonymous · Answer

let me shwo u what we have so far...im confused about teh rest

anonymous · Answer

i guesss its lnx

anonymous · Answer

after evaluating 
ln(1) = ln1 - e^-1 = 0 - 1/e = - (1/e)

ln(2) = ln2 - e^-2 = ln 2 - 1/e^2 
        1/ln 2 is less than e^2 therefore ln 2 is greater tahn 1

amistre64 · Answer

"Intermediate Value Theroem" simply says that if the ends of the interval [a,b] are opposite in sign; that the function had to cross at least once between them.|dw:1316902853167:dw|

amistre64 · Answer

it has nothing to do with greater or less than; it has to do with sign change

anonymous · Answer

my teacher gave some hint as to$$\log_{e}2 = 1 $$
so ln2 is less tahn 1

amistre64 · Answer

ok, lets try out the less than assumption|dw:1316903037056:dw|