1/5 log(0.021) = ?
take antilog of result
Find the value of (0.021)^ (1/5) using Log Tables.
5 answers
log ( a • b ) = log ( a ) + log ( b )
21 = 3 • 7
log ( 21 ) = log ( 3 • 7 ) = log ( 3 ) + log ( 7 )
0.021 = 21 / 1000
log ( a / b ) = log ( a ) - log ( b )
log ( 0.021 ) = log ( 21 ) - log ( 1000 )
If you use logarithmic tables with base 10 then:
log (1000) = 3
because 1000 = 10³
log ( 0.021 ) = log ( 21 ) - 3
log ( 0.021 ) = log ( 3 ) + log ( 7 ) - 3
log ( a ⁿ ) = n • log ( a )
log [ 0.021^ (1/5) ] = 1 / 5 log ( 0.021 ) =
[ log ( 3 ) + log ( 7 ) - 3 ] / 5
If you use logarithmic tables with base 10 then
log ( 0.021 )^ ( 1 / 5 ) = approx.=
- 0.33556 = 0.66444 - 1
Since
log ( a ) - log ( b ) = log ( a / b )
0.66444 - 1 = anti log ( 0.66444 ) / 10 =
because
1 = log ( 10 )
Result is:
4.61785 / 10 = 0.461785
If you use logarithmic tables with five decimal places, your results may differ slightly from these.
This is due to rounding logarithms on five decimal places.
21 = 3 • 7
log ( 21 ) = log ( 3 • 7 ) = log ( 3 ) + log ( 7 )
0.021 = 21 / 1000
log ( a / b ) = log ( a ) - log ( b )
log ( 0.021 ) = log ( 21 ) - log ( 1000 )
If you use logarithmic tables with base 10 then:
log (1000) = 3
because 1000 = 10³
log ( 0.021 ) = log ( 21 ) - 3
log ( 0.021 ) = log ( 3 ) + log ( 7 ) - 3
log ( a ⁿ ) = n • log ( a )
log [ 0.021^ (1/5) ] = 1 / 5 log ( 0.021 ) =
[ log ( 3 ) + log ( 7 ) - 3 ] / 5
If you use logarithmic tables with base 10 then
log ( 0.021 )^ ( 1 / 5 ) = approx.=
- 0.33556 = 0.66444 - 1
Since
log ( a ) - log ( b ) = log ( a / b )
0.66444 - 1 = anti log ( 0.66444 ) / 10 =
because
1 = log ( 10 )
Result is:
4.61785 / 10 = 0.461785
If you use logarithmic tables with five decimal places, your results may differ slightly from these.
This is due to rounding logarithms on five decimal places.
This brings back memories of teaching this 60 years ago
to find (.021)^(1/5) , this was the process:
.021 = 2.1 * 10^-2
At this point we would go to our table and find the mantissa for 2.1
which would be .3222 (I cheated and used my calculator, nobody has
log tables sitting around any more)
the characteristic would be -2
So then add (-2 + .3222) to get -1.6778
take 1/5 of that ---> -1.6778/5 = -.33556
but the mantissa has to be positive, so -.33556 = -1 + .66444
now go to the anti-log tables and "find" .6644 within the main body of the
antilog table. In most cases that number would of course not show up
exactly, so we had to calculate using difference columns
This is the stage where many students made one of many errors
(I have skipped this step and used my calculator result)
.6644 would correspond with 10^.6644 and the result is 4.6179
but the characteristic of -1 means we have to multiply this by 10^-1
so (.021)^(1/5) = 4.6179 * 10^-1 = .46179
Didn't we have fun back then ????
to find (.021)^(1/5) , this was the process:
.021 = 2.1 * 10^-2
At this point we would go to our table and find the mantissa for 2.1
which would be .3222 (I cheated and used my calculator, nobody has
log tables sitting around any more)
the characteristic would be -2
So then add (-2 + .3222) to get -1.6778
take 1/5 of that ---> -1.6778/5 = -.33556
but the mantissa has to be positive, so -.33556 = -1 + .66444
now go to the anti-log tables and "find" .6644 within the main body of the
antilog table. In most cases that number would of course not show up
exactly, so we had to calculate using difference columns
This is the stage where many students made one of many errors
(I have skipped this step and used my calculator result)
.6644 would correspond with 10^.6644 and the result is 4.6179
but the characteristic of -1 means we have to multiply this by 10^-1
so (.021)^(1/5) = 4.6179 * 10^-1 = .46179
Didn't we have fun back then ????
oh, yeah. Of course, if 3 digits was enough, you always reached for the slide rule instead.
I remember our library had huge volumes of 16-place log and trig tables, published by the US government.
I remember our library had huge volumes of 16-place log and trig tables, published by the US government.
Just curious here, where are they still teaching how to use log tables??
1 answer