Question

Can anyone teach me how to solve tough calculations using logs?

Answer 1

Here is a link to Khan's videos
http://www.khanacademy.org/math/algebra/logarithms

See if they help.

Answer 2

that is not what i want..i want something like this:
if i have a multiplication like
23982104*28427194827
we take logs on both sides..from the log table
then take anti logs etc..
i cant find the tut to that anywhere...i know basic logs

Answer 3

you are asking about "obsolete" technology...
The first step is you need a table of logs (base 10) (though we could use base e)

Do you have such a table?

Answer 4

yes

Answer 5

the first step is write your problem in scientific notation
23982104 --> 2.3982e7 (means 2.3982 times 10^7)
28427194827 --> 2.8427e10
we only keep as many digits as your table supports. How many digits does it support?

Answer 6

5 or so

Answer 7

I have a table that shows 3 digits going down the page, a 4th digit going across, and you can use interpolation to get a 5th digit

Answer 8

for 2.3982e7
looking at this table I see log10(239) to 5 digits is
37840
log10(2398) is 37985 and log10(2399) is 38003 (18 higher)
we can interpolate log10(23982) to be 37989

that means 10^0.37989 is about 2.3982
type
10^0.37989=
in the google search window to verify
to account for the factor of 10^7 we prepend a 7:

log10(2.3982e7) = 7.37989
check with google
10^7.37989=
it gives us 23982254.1 which is close to the original number.

use a similar procedure to find the log base 10 of
2.8427e10 is 10.45373

multiplying two numbers becomes adding their logarithms:
7.37989+ 10.45373= 17.83362

now find the antilog
this is 10^17 times 10^0.83362
Look in the table at the logarithms for 0.83362 the nearest entry in my table is
83359 which corresponds to 6817. we want 83362 or 3 higher. this corresponds to 68174

so the answer is
6.8174e17

you can use a calculator or google to check this result.