# does anyone know wht this means?

• Nov 15th 2007, 05:00 PM
lmao6
does anyone know wht this means?
Hey, im doing data management for a engineering lab, and one of the questions asks to prove that doing the semi-log plot of a graph with exponential correlation give the same relation. A semi-log graph is taking all the y values of the exponential graph and applying the natural logirthm to it.
ex: y= 6 in the semi long graph y= ln 6
This will give a linear relationship which is easier to work with.
My question is, they ask me to prove this, by taking the regression equation algebraically alter it to equal thre regression equation of the linear relation ship
this isthe exponential relation:

1E-230e^(0.2682x)
What exactly does the E mean, and how can i manipulate this to equal,

0.1165x – 229.96
• Nov 15th 2007, 05:05 PM
Jhevon
Quote:

Originally Posted by lmao6
Hey, im doing data management for a engineering lab, and one of the questions asks to prove that doing the semi-log plot of a graph with exponential correlation give the same relation. A semi-log graph is taking all the y values of the exponential graph and applying the natural logirthm to it.
ex: y= 6 in the semi long graph y= ln 6
This will give a linear relationship which is easier to work with.
My question is, they ask me to prove this, by taking the regression equation algebraically alter it to equal thre regression equation of the linear relation ship
this isthe exponential relation:

1E-230e^(0.2682x)
What exactly does the E mean, and how can i manipulate this to equal,

0.1165x – 229.96

i'm pretty sure the E means 10^, that's what it means on my calculator, for instance.

if i enter $\displaystyle 1.2 \times 10^2$, then $\displaystyle 1.2 E2$ shows up on my display screen
• Nov 15th 2007, 05:09 PM
lmao6
thank you SO much, that was bugging me ther whole day.
I can finally solve this question:D many thanks
btw im new to these forums, but i think i will be putting alot more time in here.
• Nov 15th 2007, 05:12 PM
Jhevon
Quote:

Originally Posted by lmao6
thank you SO much, that was bugging me ther whole day.I can finally solve this question:D many thanks

you're welcome

Quote:

btw im new to these forums, but i think i will be putting alot more time in here.
ok. good
• Nov 15th 2007, 05:21 PM
lmao6
heh back for a bit more help...
so this is what i tried:
log(10^(-230e^(0.2682x))
=-230e^(0.2682x)(1)
=ln(-230e^(0.2682x))
=ln (-230) + ln e^(0.2682x)
=ln (-230) + 0.2682x
almost got it! but im not sure what im doing wrong :S
0.1165x – 229.96 is what im trying to get