1. ## 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?

2. 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
...

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

4. 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.
.

5. 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

6. 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
$\displaystyle 60.52 = 19.05 \ln(x)$

divide by $\displaystyle 19.05$ ...

$\displaystyle \frac{60.52}{19.05} = \ln(x)$

convert the log equation into an exponential equation ...

$\displaystyle e^{\frac{60.52}{19.05}} = x$

pop this value into your calculator ...

$\displaystyle x = 23.97...$

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

8. 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.