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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ln(2x-5)=0.5
2x-5=e^0.5
2x=e^0.5+5
x=(e^0.5+5) / 2
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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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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e=mc2

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e is a log of x where the gradient = the y value

and ln x is the inverse of e in y=x
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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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,
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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e=mc2

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E=mc²
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