#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

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

#2

do your own god damn homework!

#4

ln(2x-5)=0.5

2x-5=e^0.5

2x=e^0.5+5

x=(e^0.5+5) / 2

2x-5=e^0.5

2x=e^0.5+5

x=(e^0.5+5) / 2

#5

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

#6

ln(2x-5)=0.5

2x-5=e^0.5

2x=e^0.5+5

x=(e^0.5+5) / 2

This.

#7

e=mc2

#8

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

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

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

#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

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

#10

OMG i was right!!! thanks for the wikipage aswell btw! Now i actually know what the hell e and ln are.

*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.

so instead of the normal log base 10, you substitute e for 10, and instead of writing log,you write ln.

#12

the maths thread can.

#13

e=mc2

Superscript fail .

E=mc²