Thread: logarithm regression equation

The table shows Taiwan's nuclear power generation data in billions of killowatt hours. Let x=5 represent 1980, x=10 rep. 1985 and so on.

Year Energy produced
1980 7.8
1985 27.8
1990 31.6
1995 33.9


1) find natural logarithm regression equation for the data.
y=-20.52 + 19.05 ln(x)

I got this in my calculator..is it correct? Thanks!

2) Predict when the nuclear power generation will reach 40 billion kilowatt hours.
I have a question -- do you plug 40 billion into the x or the y? I think y, because it is total power, but just checking my work.

When I plug 40 billion in for y, I come up with 21.47, which would be year 1996.
Am I correct?

Originally Posted by live_laugh_luv27

The table shows Taiwan's nuclear power generation data in billions of killowatt hours. Let x=5 represent 1980, x=10 rep. 1985 and so on.

Year Energy produced
1980 7.8
1985 27.8
1990 31.6
1995 33.9


1) find natural logarithm regression equation for the data.
y=-20.52 + 19.05 ln(x)

ok

2) Predict when the nuclear power generation will reach 40 billion kilowatt hours.
I have a question -- do you plug 40 billion into the x or the y? I think y, because it is total power, but just checking my work.

When I plug 40 billion in for y, I come up with 21.47, which would be year 1996.
Am I correct?

y = 40 ... x = 23.97 ... about 1999

...

would you plug in 40,000,000,000 or 40? because it's already in billions of kilowatts?

Originally Posted by live_laugh_luv27

would you plug in 40,000,000,000 or 40? because it's already in billions of kilowatts?

The table shows Taiwan's nuclear power generation data in billions of killowatt hours.

.

Originally Posted by skeeter

.

ok, when I plug in 40, I'm stuck at 60.52 = 19.05ln(x).
When I try to divide to get rid of the ln(x) I'm coming up with completely wrong answers..help? Thanks

Originally Posted by live_laugh_luv27

ok, when I plug in 40, I'm stuck at 60.52 = 19.05ln(x).
When I try to divide to get rid of the ln(x) I'm coming up with completely wrong answers..help? Thanks

divide by ...



convert the log equation into an exponential equation ...



pop this value into your calculator ...

ok, thanks...so the end of 1998. or should I round it to 1999?

Originally Posted by live_laugh_luv27

ok, thanks...so the end of 1998. or should I round it to 1999?

I'm not going to hold your hand on this ... you decide.