Finding the Altitude at a Given Air Pressure

[tardisblue]

Atmospheric pressure decreases as altitude increases. The pressure can be approximated by the formula \(\displaystyle P(h) = 101.33(0.9639)^\frac{h}{1000}\) where \(\displaystyle P\) is pressure in kilopascals and \(\displaystyle h\) is height in feet.
a) What is the pressure at sea level (0 feet)?
b) At what altitude will the air pressure be 70kPa?So, I kinda have an idea how to do this. For a), you just sub in 0 for h, right? And that gets you 101.33

For b), do you just do trial and error until you get it to equal 70 feet? Could someone show the process/if there's an easier way?

Thanks!

 

[MarkFL]

For part b) you can use logarithms to solve for $h$. Let $P(h)=70$:

\(\displaystyle 70=101.33(0.9639)^{\frac{h}{1000}}\)

Now, if you have an equation of the form:

\(\displaystyle a=b\cdot c^{\frac{x}{d}}\)

and you wish to solve for $x$, the first thing I would do is divide through by $b$:

\(\displaystyle \frac{a}{b}=c^{\frac{x}{d}}\)

Next, we may take the natural log of both sides:

\(\displaystyle \ln\left(\frac{a}{b}\right)=\ln\left(c^{\frac{x}{d}}\right)\)

On the right, we may use the log property \(\displaystyle \log_a\left(b^c\right)=c\cdot\log_a(b)\) to write:

\(\displaystyle \ln\left(\frac{a}{b}\right)=\frac{x}{d}\ln(c)\)

Next divide through by \(\displaystyle \frac{\ln(c)}{d}\):

\(\displaystyle \frac{d\ln\left(\frac{a}{b}\right)}{\ln(c)}=x\)

Can you apply this procedure to the given problem?

 

[tardisblue]

Thanks for replying!
We don't go over logarithms for this class (that's next year), so I don't think that my teacher would want us to solve it like that, since this is a question on a test review.

So, is there any other method of solving that, without logarithms? I got the answer \(\displaystyle 10\) kPa through trial and error. Is that correct? Thanks again! (: