MHB Finding the Altitude at a Given Air Pressure

[tardisblue]

Atmospheric pressure decreases as altitude increases. The pressure can be approximated by the formula $$P(h) = 101.33(0.9639)^\frac{h}{1000}$$ where $$P$$ is pressure in kilopascals and $$h$$ 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 $P(h)=70$:

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

Now, if you have an equation of the form:

$$a=b\cdot c^{\frac{x}{d}}$$

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

$$\frac{a}{b}=c^{\frac{x}{d}}$$

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

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

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

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

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

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

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 $$10$$ kPa through trial and error. Is that correct? Thanks again! (: