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Calculation of logs

hp12345

New member
Aug 6, 2012
10
Hi,
I have been solving pH problems since last 2 days but the problem is that I don't know any method to calculate logarithms without using calculators,so anybody can spare few minutes to show me how to calculate(estimate in general sense) common logarithms ?


[to mod-I wasn't sure where to put that so please move it if thread is in the wrong place :)]
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Hi,
I have been solving pH problems since last 2 days but the problem is that I don't know any method to calculate logarithms without using calculators,so anybody can spare few minutes to show me how to calculate(estimate in general sense) common logarithms ?


[to mod-I wasn't sure where to put that so please move it if thread is in the wrong place :)]
You can in principle calculate logs by hand, but it is not really practical. Before we had cheap calculators we used log tables and slide rules.

But if you only need rough values you can use the following common (base 10) logarithms \(\log(1)=0\), \(\log(10)=1\), \(\log(100)=2\) etc. \(\log(2)\approx 0.3\), \(\log(2.5)\approx 0.4\), \(\log(4)\approx 0.6\) \(\log(5)\approx 0.7\), \(\log(6.25)\approx 0.8\), \(\log(8)\approx 0.9\) and the laws of logarithms to estimate other logs base 10.

CB
 
Last edited:

hp12345

New member
Aug 6, 2012
10
Thank you for replying.
I think I have got it,I will come back if I will need help :)
 

hp12345

New member
Aug 6, 2012
10
Hi,
So now I am quite used to estimate pH but now the problem arises when it comes to antilogs.
For example a question states that-ph of a solution is 3.28 so find its H+ ion concentration.
As I know pH = - log[H+],on solving by taking antilog I am left with [H+] =10^-3.28 which I further simplifies to [H+]=10^-3*10^-0.28 so the question is how to calculate 10^-0.28 without calculator ?
or in short how to calculate decimal exponents ?

I referred many books,vids and net but everywhere nobody shows the way to calculate that without calculator. :confused:
 

earboth

Active member
Jan 30, 2012
74
Hi,
So now I am quite used to estimate pH but now the problem arises when it comes to antilogs.
For example a question states that-ph of a solution is 3.28 so find its H+ ion concentration.
As I know pH = - log[H+],on solving by taking antilog I am left with [H+] =10^-3.28 which I further simplifies to [H+]=10^-3*10^-0.28 so the question is how to calculate 10^-0.28 without calculator ?
or in short how to calculate decimal exponents ?

I referred many books,vids and net but everywhere nobody shows the way to calculate that without calculator. :confused:
1. Transform the value -3.28 into a sum of an integer and a positive decimal fraction. You'll get:

$-3.28 \approx -4 + 0.7$

2. Now use base 10:

$10^{-3.28} \approx 10^{-4+0.7} = 10^{-4} \cdot 10^{0.7}$

3. Now use the table posted by CB:

$10^{-3.28} \approx 10^{-4+0.7} = 10^{-4} \cdot 10^{0.7} = 0.0001 \cdot 5 = 0.0005$

If you use a calculator (for confrimation only!) you'll get: $10^{-3.28} \approx 0.000525$


btw: If you have a new problem please start a new thread.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
I might as well complete the little table of approximate logs I started in an earlier post:

\[\begin{array}{|c|c|} \hline \\ x & \log_{10}(x)\\ \hline \\ 1.25& 0.1 \\ 1.56 & 0.2 \\ 2 & 0.3 \\ 2.5 & 0.4 \\ 3.125 & 0.5 \\
4 & 0.6 \\ 5 & 0.7 \\ 6.25 & 0.8 \\ 8 & 0.9 \\ \hline \end{array} \]

Note you can produce this table yourself just by knowing that \(\log_{10}(2)\approx 0.3\) and using the laws of logarithms. The relative errors are all below 5%, and other than the first two better than 1.1%

CB
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I have used the well-known approximation $\displaystyle 2^{10}\approx10^3$ to show students how to obtain $\displaystyle \log(2)\approx0.3$, and then the properties of logs to get other approximations as suggested by CB.
 

hp12345

New member
Aug 6, 2012
10
btw: If you have a new problem please start a new thread.
err,sorry for that,I will keep that in mind next time.
btw thank you earboth and cb (Smile)