- #1
cbarker1
Gold Member
MHB
- 349
- 23
Hi everyone,
I need some help to solve this problem:
The direction states to find the value by using the log table
$\log_{3}\left({825.6}\right)$
Work:
I using the change of base:
$\log_{3}\left({825.6}\right)=\frac{\log\left({825.6}\right)}{\log\left({3}\right)}$
I look up the values of the logarithm of 3 and 825.6.
$\log\left({825.6}\right)=.91677+2$ =$\log\left({8.256}\right)+2$
$\log\left({3}\right)=.47712$
yields
$\log_{3}\left({825.6}\right)=\frac{2.91677}{.47712}$
Taking the log of both sides:
$\log\left({\log_{3}\left({825.6}\right)}\right)=\log\left({\frac{2.91677}{.47712}}\right)$
Using the log identity to dividing to difference:
$\log\left({\log_{3}\left({825.6}\right)}\right)=\log\left({2.91677}\right)-\log\left({.47712}\right)$
Finding the values of log(2.91677) and log(.47712)
Do I drop the last digit in (2.91677 and .47712) and look up the first four digits or do I interpolate the last digits by using the Table of proportional parts?
Thank you,
Cbarker
I need some help to solve this problem:
The direction states to find the value by using the log table
$\log_{3}\left({825.6}\right)$
Work:
I using the change of base:
$\log_{3}\left({825.6}\right)=\frac{\log\left({825.6}\right)}{\log\left({3}\right)}$
I look up the values of the logarithm of 3 and 825.6.
$\log\left({825.6}\right)=.91677+2$ =$\log\left({8.256}\right)+2$
$\log\left({3}\right)=.47712$
yields
$\log_{3}\left({825.6}\right)=\frac{2.91677}{.47712}$
Taking the log of both sides:
$\log\left({\log_{3}\left({825.6}\right)}\right)=\log\left({\frac{2.91677}{.47712}}\right)$
Using the log identity to dividing to difference:
$\log\left({\log_{3}\left({825.6}\right)}\right)=\log\left({2.91677}\right)-\log\left({.47712}\right)$
Finding the values of log(2.91677) and log(.47712)
Do I drop the last digit in (2.91677 and .47712) and look up the first four digits or do I interpolate the last digits by using the Table of proportional parts?
Thank you,
Cbarker
Last edited: