- #1
gerid21
- 2
- 0
- Homework Statement
- Calculate ##890\times12.34\times0.0637## using logarithms of base 10
- Relevant Equations
- Log tables, calculator
From the log tables:
##log(890) = 2.9494, \space
log(12.34)=1.0913, \space
log(0.0637)=\bar{2}.8041##
I calculate by hand:
##\begin{array}{r}
&2.9494\\
+&1.0913\\&\bar{2}.8041\\\hline &2.8448
\end{array}##
Thus:
##log^{-1}(2.8448) \approx 699.6 \space##
Which is the correct answer.
Now I know ##log(0.0637)=\bar{2}.8041=-2+0.8041=-1.1959##
Yet if I did
##
\begin{array}{r}
&2.9494\\
+&1.0913\\&-1.1959\\\hline &3.2366
\end{array}##
##log^{-1}(3.2366)\approx1724\space## This is obviously wrong.
But if I were to do the problem solely with a calculator, I have to plug in ##log(0.0637)= -1.1959\space## and not ##-2.8041## to get the correct answer. I'm confused at what's going on. I know it has something to do with the negative characteristic of the logarithm. But then if I'm adding it by hand, why doesn't ##-1.1959## work?
##log(890) = 2.9494, \space
log(12.34)=1.0913, \space
log(0.0637)=\bar{2}.8041##
I calculate by hand:
##\begin{array}{r}
&2.9494\\
+&1.0913\\&\bar{2}.8041\\\hline &2.8448
\end{array}##
Thus:
##log^{-1}(2.8448) \approx 699.6 \space##
Which is the correct answer.
Now I know ##log(0.0637)=\bar{2}.8041=-2+0.8041=-1.1959##
Yet if I did
##
\begin{array}{r}
&2.9494\\
+&1.0913\\&-1.1959\\\hline &3.2366
\end{array}##
##log^{-1}(3.2366)\approx1724\space## This is obviously wrong.
But if I were to do the problem solely with a calculator, I have to plug in ##log(0.0637)= -1.1959\space## and not ##-2.8041## to get the correct answer. I'm confused at what's going on. I know it has something to do with the negative characteristic of the logarithm. But then if I'm adding it by hand, why doesn't ##-1.1959## work?
Last edited: