Finding $n^{th}$ root using logarithms

Question

The following is the question I'm stuck at:

Find the seventh root of 0.00324, having given that $$\log 44092388 = 7.6443636$$

Now my approach was as follows: Let $$x=(0.00324)^\frac {1}{7}$$ $$\Rightarrow \frac {1}{7}( \log 324 -5)=\log x$$ But since I have been given only $\log 44092388 = 7.6443636$ is it possible that I can find the logarithm without using logarithms table to find the $\log 324$?

Answer 1

I'm going to go out on a limb and say no, you need to check log tables twice to solve this problem.

Here's how we did it in my day, when log tables were all we had. $\log 3.24=0.5145$, so we would write $$0.00324=3.24\cdot10^{-3}=3.24\cdot10^{67-70}\\\log0.00324=67.5105-70\\\frac17\log0.00324=9.6443-10\\\sqrt[7]{0.00324}=10^{9.6443-10}=4.41\cdot10^{-1}=0.441$$

(All of these equalities, of course, should be approximations. Also, in my day, we didn't assume more significant digits that we had in the original number.)

Answer 2

You can write $$\ln(x)=\frac{1}{7}\left(\ln(171)-\ln(500000)\right)$$

Answer 3

Manipulating the given equality we have:

$$\log_{10} 44,092,388 - \log_{10} 100,000,000 = 7.6443636 - 8$$ $$\log_{10} 0.44092388 = -0.3556364$$ $$0.44092388 = 10^{-0.3556364}$$ $$0.00324^{1/7} = 0.440924$$

Of course, you need to first realise $0.00324^{1/7} = 0.44092388 \cdots$ as mentioned by J. W. Tanner. From just looking at the question, there seems to be no relation between these two numbers, so I would say the question is quite poorly constructed.