Asked by Anonymous
Use log table to solve these;
2.647*0.00921
/
0.05738
2.647*0.00921
/
0.05738
Answers
Answered by
Reiny
Don't tell me that you are still using log "tables" ...
(The last time I taught how to use log tables must be more than 40 years ago.)
I no longer have any texts that include log tables, so on line here is one I found and will use:
http://myhandbook.info/table_commonlog.html
I also have to remember all that mantissa and characteristic stuff.
let x = 2.647*0.00921 / 0.05738
log x = log (2.647*0.00921 / 0.05738)
= log 2.647 + log 0.00921 - log 0.05738
= log 2.647 + log(9.21 x 10^-3) - log (5.738 x 10^-2)
= log 2.647 + log 9.21 -3 - log 5.738 + 2
now using the tables
= .422754 + .964260 - .758761 - 1
= .628253 -1
At this point I would have gone to an "antilog" set of tables,
however I will no try to find .628253 in our table from above
x = antilog .628253 x 10^-1
= between 4.248 and 4.249 x 10^-1
how about x = .42485
actual calculated answer = .42487
(The last time I taught how to use log tables must be more than 40 years ago.)
I no longer have any texts that include log tables, so on line here is one I found and will use:
http://myhandbook.info/table_commonlog.html
I also have to remember all that mantissa and characteristic stuff.
let x = 2.647*0.00921 / 0.05738
log x = log (2.647*0.00921 / 0.05738)
= log 2.647 + log 0.00921 - log 0.05738
= log 2.647 + log(9.21 x 10^-3) - log (5.738 x 10^-2)
= log 2.647 + log 9.21 -3 - log 5.738 + 2
now using the tables
= .422754 + .964260 - .758761 - 1
= .628253 -1
At this point I would have gone to an "antilog" set of tables,
however I will no try to find .628253 in our table from above
x = antilog .628253 x 10^-1
= between 4.248 and 4.249 x 10^-1
how about x = .42485
actual calculated answer = .42487
Answered by
Anita
I don't understand
Answered by
Peace
I don't understand
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