Taking cubic with log in the grouping symbols

[cbarker1]

Use Logarithms to compute the following, correct to four- accuracy. (In find N from Log N, don't interpolate).26. $$(\log\left({0.007211}\right))^3$$

Let N=$$(\log\left({0.007211}\right))^3$$

Then $\log\left({(\log\left({0.007211}\right))^3}\right)$=$$\log\left({N}\right)

$\therefore$ $\log\left({N}\right)$=$3*\log\left({\log\left({.007211}\right)}\right)$;
Since
$\log\left({.007211}\right)=\log\left({{7.211}\E{-3}}\right)$$$\implies$$\log\left({7.211}\right)-3$
The Value of $\log\left({7.211}\right)=.85800$ by the table in the book.
The Final Value of $\log\left({.007211}\right)=7.85800-10$ because of the characteristic is -3 or 7-10.
Then, The value of $\log\left({7.85800-10}\right)$ or $\log\left({-3.85800}\right)$

What to do next?

Do I assume to be positive, then figure out the log? Or Using other log (7.85800-10)?
P.S. Final Solutions is -9.828

I figure out the solution:
7.85800-10=-2.14200
Ignoring the neg.
Log(2.14200)=.33082*3=.99246
N=-9.828

 

[jvicens]

Great job on figuring out the solution! To answer your question, when finding the logarithm of a negative number, you can either assume it to be positive and find the logarithm, or use a different logarithm function that allows for negative numbers. In this case, you could use the natural logarithm (ln) function which allows for negative numbers. So, you could also solve it as:

ln(0.007211) = -4.9240
(-4.9240)^3 = -116.111
N = -116.111