SOLUTION: The altitude of an aircraft, h, in miles, is given by h= - (100/9)log p/b where P = the outside pressure and B = the atmospheric pressure at sea level. Let B = 31 inches of mer

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: The altitude of an aircraft, h, in miles, is given by h= - (100/9)log p/b where P = the outside pressure and B = the atmospheric pressure at sea level. Let B = 31 inches of mer      Log On


   

Question 713101: The altitude of an aircraft, h, in miles, is given by
h= - (100/9)log p/b
where P = the outside pressure and B = the atmospheric pressure at sea level. Let B = 31 inches of mercury. What is the outside air pressure at a height of 2.7 miles? Round your answer to the nearest tenth.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
h=+-+%28100%2F9%29log%28%28p%2Fb%29%29
Substituting in the values given for b and h we get:
2.7+=+-%28100%2F9%29log%28%28p%2F31%29%29
And now we solve for p. First we'll isolate the log. Multiplying both sides by -9/100:
%28-9%2F100%29%2A%282.7%29+=+%28-9%2F100%29%2A%28-%28100%2F9%29log%28%28p%2F31%29%29%29
-.243+=+log%28%28p%2F31%29%29
Next we rewrite the equation in exponential form. In general log%28a%2C+%28p%29%29+=+n is equivalent to p+=+a%5En. Using this pattern (and the fact that the base of "log" is 10) on our equation we get:
10%5E%28-.243%29+=+p%2F31
Next we multiply both sides by 31:
31%2A10%5E%28-.243%29+=+p
This is an exact expression for the solution. I'll leave it up to you to use your calculator to find the decimal. (Just be sure to raise 10 to the -.243 power before you multiply by 31.)