Question 278784: I am trying to figure out a problem. I have some of it.
I just don't know how to do the last part of it.
The end of it is
(62.35)+(-0.46)ln x =61
How do I solve this equation? The actual answer is 18.8. But, I cannot even get close to that. In class we use the TI-84. Thanks
If you need the entire problem let me know. Thanks so much.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! working from your equation of:
(62.35)+(-0.46)ln x =61
subtract 62.35 from both sides of this equation to get:
-.46*ln(x) = 61 - 62.35
simplify to get:
-.46*ln(x) = -1.35
divide both sides of this equation by -.46 to get:
ln(x) = -1.35 / -.46 = 2.934782609
use your calculator to find the number whose natural log is 2.934782609 to get:
x = 18.81741195
here's a link to instructions for the TI-84
http://education.ti.com/guidebooks/graphing/84p/TI84PlusGuidebook_Part2_EN.pdf
In my calculator (TI-30), in order to get the natural log of a number, I enter the number and then press the LN key.
If I want to get the number whose natural log is a number, I enter the number and then press the 2D key and then press the LN key.
That's shown as the e^x key in gold letters on top of the LN key.
Looks like you have your LN key to the left of the 4, and looks like you have e^x key being in the 2d function on that same key.
Do the following:
Enter 10 and then hit the LN key.
you should see 2.302585093
hit the 2d LN key.
you should see your 10 come back.
ln(x) and e(x) are inverse functions.
y = ln(x) if and only if x = e^y
your x was 10
your y was 2.302585093
basic definition becomes:
2.302585093 = ln(10) if and only if e^(2.302585093) = 10 which it does.
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