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 $\log 324$?

I don't see any obvious connection between $44092388$ and $324$. — lulu, Oct 27 '19 at 11:56
@J.W.Tanner Sure, but would someone spot this without actually solving the problem first? — lulu, Oct 27 '19 at 11:59

Toby Mak · Answer 1 · 2019-10-27T12:29:56.860

1

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.

edited Oct 27 '19 at 12:29

answered Oct 27 '19 at 12:16

Toby Mak

16,827

Sorry, just realised there were a couple of mistakes in the post. – Toby Mak Oct 27 '19 at 12:20
@MatthewDaly: what you wrote in your comment is not correct; what did you mean? – J. W. Tanner Oct 27 '19 at 12:24
That comment was in response to a previous version of my answer which was incorrect. I assume Matthew means that: 'you need to recognise that $10^{-0.3556364} = 0.440924$'. – Toby Mak Oct 27 '19 at 12:25
@J.W.Tanner My point is that you can manipulate the log fact that was given to you all you like, but you also have to do some math or have an additional log fact that ties any of it to $0.00324$. – Oct 27 '19 at 12:28

score 1 · Accepted Answer · 2019-10-27T12:31:54.517

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.)

score 0 · Answer 3 · answered Oct 27 '19 at 11:55

0

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

answered Oct 27 '19 at 11:55

Dr. Sonnhard Graubner

95,283

Finding $n^{th}$ root using logarithms

3 Answers3