Algebra 2 natural logarithms problem

[mustang]

Problem 51.
At a constant temperature, the atmospheric pressure p, in pascals, is given by the formula p=101.3e^-0.001h, where h is the altitude in meters. Find h when p is 74.3 pascals. Hint: Start by taking the natural logarithms of the expressions on each side of the equation.

Solve and check.
41. (29.3)^4x-1=(17.3)^5x+2

 

[cookiemonster]

1. The problem tells you how to solve it. What've you tried?

2. Manipulate the expression to get a common exponent, and then take the logarithm. What've you tried so far?

cookiemonster

 

[mustang]

regard on problem 51.

So it would be:
p=74.3
p=101.3e^-0.001h
74.3=101.3e^-0.001h
74.3=274.3619^-0.001
Is this right, so far? If so what is the next step??

 

[enigma]

74.3=101.3e^-0.001h
74.3=274.3619^-0.001

What did you do to go through that step?

a latural logarithm is like "the opposite" of an e^

If you put them together, you get 1.

So, take ln of both sides to get rid of the e^-0.001*h, and give you *some number* times -.001h on the right side, and ln(74.3) on the left.

 

[cookiemonster]

What'd you do between the third and fourth line? How'd you make the h disappear? Why wasn't the first step you did taking the natural logarithm of both sides like the problem says?

cookiemonster

 

[mustang]

Cont...

p=74.3
p=101.3e^-0.001h
74.3=101.3e^-0.001h
ln(74.30)=ln(101.3)*-0.001h
4.3081=4.618086*-0.001h
Is this right so far??

Any given program, when running, is obsolete.
(Laws of Computer Programming, I)

 

[cookiemonster]

Careful.

ln(101.3*e^(-.001h)) = ln(101.3) + ln(e^(-.001h)) = ln(101.3) - .001h

cookiemonster