#1
ok basically we've got some homework and it has a question on e and ln, but I havent covered these in class. So can someone help me with the questions below and explan along the way? thanls guys

oh yeah ive learnt that e is a constant with applications in compound interest using wikipedia and that e^ln(0.5)=0.5 using my calculator.

ln(2x-5)=0.5 (i tried with my limited knowledge and got (e^(0.5)+5)/2 but i think its worng)
and
e^(2x-5)=0.5
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#5
Quote by SmarterChild
ln(2x-5)=0.5
2x-5=e^0.5
2x=e^0.5+5
x=(e^0.5+5) / 2

You need to clean it up a bit. All your arithmetic is correct, it's just not notated correctly.

2x-5=e^0.5
2x=(e^0.5)+5
x=[(e^0.5)+5] / 2
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#6
Quote by SmarterChild
ln(2x-5)=0.5
2x-5=e^0.5
2x=e^0.5+5
x=(e^0.5+5) / 2


This.
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#7
e=mc2

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#9
e^(2x+5)=0.5
ln(e^(2x+5)=ln(0.5) <-- ln & e cancel out
2x+5=ln(0.5)
2x=ln(0.5) -5
x=(ln(0.5) -5) /2
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#10
OMG i was right!!! thanks for the wikipage aswell btw! Now i actually know what the hell e and ln are.
"Like a midget at a urinal, I was going to have to stay on my toes"

"Like a blind man at an orgy, I was going to have to feel my way through"

Last edited by YngwieJnr at Nov 6, 2008,
#11
ln is log base e.

so instead of the normal log base 10, you substitute e for 10, and instead of writing log,you write ln.
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#12
the maths thread can.
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#13
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e=mc2


Superscript fail .

E=mc²
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