It's been a good few years since I've had to touch logs with my job, however today I found myself trying to calculate 570=a(1.5)^n for a theoretical pressure system.

Something in my head from high school tells me that I need to use logs to solve for a and n. Admittedly I could do it using a graph but I'd quite like to be able to do logs again as I know they are useful.

So onto the question. Just how do I utilise logs to solve 570=a(1.5)^n for a and n.

So far I have used my basic knowledge and taken logs of both sides (possibly not correctly) to get -


Then I rearranged to get -


This is where i'm stuck. I haven't got 'a' so I can't do anything more at this point, can I?

The basic formula I am using is T=ap^n. My values for T and p are 570 and 1.5 respectively.


I have been told to work it out using T=570 when p=1.5 AND T=510 when p=1. I am an idiot and pieces of paper have two sides...

However if I implement this into my equation as above, I am still confused.I now have -

n=(log510)/(loga) and n=(log570)/(log1.5a)

Am I missing something here?

  • $\begingroup$ That is the simplest you can get. $\endgroup$ – Parcly Taxel Nov 2 '16 at 12:52
  • $\begingroup$ @ParclyTaxel The formula I'm using is T=ap^n. Could this be 'logged' to find a and n? $\endgroup$ – Ctrl-alt-dlt Nov 2 '16 at 12:56
  • $\begingroup$ But there are two variables! You may find $a$ in terms of $n$ or vice versa, but nothing more. $\endgroup$ – Parcly Taxel Nov 2 '16 at 12:56

You have $570=a*1.5^n$ and $510=a*1^n$. Since $1^n=1$ for all $n$ you have $a=510$. Rewriting the first equation you have:

$570=510*1.5^n$ Which simplifies to $\frac{19}{17}=1.5^n$. Using logs we find $n=\frac{ln(19)-ln(17)}{ln(1.5)} =.2743$ish. So $a=510$, $n=.2743$


You don't.

You have two variables and just one equation, so there would be an $a$ for every $n$ in your system. If you want to solve it, you would need one more equation.

----------------------------After EDIT----------------------------

So, basically, you have got some stuff wrong. Consider - $T = a(p)^n \implies \log T = \log a + n\log(p)$

Thus, your equations become

$\log 570 = \log a + n\log(1.5), \log 510 = \log a + n \log 1$.

Solving them would give you, $a = 510$ and $n = \log(570/510) / \log(1.5)$

  • $\begingroup$ So my boss is wasting my time trying to work this thing out then, haha. Thanks for your response :) $\endgroup$ – Ctrl-alt-dlt Nov 2 '16 at 12:58
  • $\begingroup$ If you want to get over with it, hand him a graph b/w $a$ and $n$ highlighting some values. That way, you would end the conversation on a positive note. $\endgroup$ – Pratyush Rathore Nov 2 '16 at 13:02
  • $\begingroup$ Please be aware that I have updated my question due to my own incompetence. $\endgroup$ – Ctrl-alt-dlt Nov 2 '16 at 13:07

$T=ap^n$ linearizes by the use of logarithms:

$$n\log p+\log a=\log T.$$

Then with the given data, you get a linear system of two equations, in the unknowns $\log n$ and $n$.

$$n\log1.5+\log a=\log570,\\ n\log1+\log a=\log510.$$

As $\log1=0$, the second equation immediately yields $a=510$, and the first gives $$n=\frac{\log570-\log510}{\log1.5}.$$

In a more general case with $(p_1,T_1),(p_2,T_2)$ you could solve the system as

$$n=\frac{\log T_2-\log T_1}{\log p_2-\log p_1},\\ a=\frac{T_1}{p_1^n}.$$


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