Log Question

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February 28th 2008, 11:54 AM
bobak

Log Question

Quote:

To nine decimal places, $\log_{10}{2} = 0.301029996$ and $\log_{10}{3} = 0.477121255$

i) Calculate $\log_{10}{5}$ and $\log_{10}{6}$ to three decimal places. By taking logs, or otherwise, show that

$5 \times 10^{47} < 3^{100} < 6 \times 10^{47}$
Hence write down the first digit of $3^{100}$

ii) Find the first digit of each of the following numbers: $2^{1000}$ , $2^{10 \ 000}$ and $2^{100 \ 000}$

[OCR STEP I 2000, Question 1]

I could do with a tiny bit of guidance on this. thanks in advance for any help

- Bobak
February 28th 2008, 12:52 PM
Soroban

Hello, bobak!

Here's the first one . . .

Quote:

Given: . $\begin{array}{ccc}\log_{10}2 &=&0.30103 \\ \log_{10}3 &=&0.47712 \end{array}$

calculate $(a)\;\log_{10}5\text{ and }(b)\;\log_{10}6$ to three decimal places.

$(a)\;\;\log_{10}(5) \;=\;\log_{10}\left(\frac{10}{2}\right) \;=\;\log_{10}(10) \;- \;\log_{10}(2) \;=\;1 \;- \;0.30103 \;\approx\;0.699$

$(b)\;\;\log_{10}(6) \;=\;\log_{10}(2\cdot3) \;=\;\log_{10}(2)\;+ \;\log_{10}(3) \;=\;0.30103 \;+\;0.47712 \;\approx \;0.778$
February 28th 2008, 01:54 PM
mr fantastic

Quote:

Originally Posted by bobak

I could do with a tiny bit of guidance on this. thanks in advance for any help

- Bobak

And here's the next:

It boils down to showing $47 + \log_{10} 5 < 100 \log_{10} 3 < 47 + \log_{10} 6$ .

And you know the values of all the logs from the earlier part ......

So clearly $3^{100} = 5.abcd..... \times 10^{47}$ - its first digit is now obvious.
February 28th 2008, 02:04 PM
mr fantastic

Quote:

Originally Posted by bobak

I could do with a tiny bit of guidance on this. thanks in advance for any help

- Bobak

For the final bit, try to apply the reasoning from earlier:

Can it be shown that $a \times 10^x < 2^{1000} < (a + 1) \times 10^x$ , where obviously a and x aren't known yet!

Note that $\log_{10} 2^{1000} = 1000 \log_{10} 2 = 301.029996$ . So clearly x = 301.

So can you find an a such that $a \times 10^{301} < 2^{1000} < (a + 1) \times 10^{301}$ . In other words, can you find an a such that

$<br /> 301 + \log_{10} a < 301.029996 < 301 + \log_{10} (a + 1)$ .

Again, you'd use the known values of log from earlier. Warning: spoiler below.

Answer: a = 1 so the first digit is 1.
February 28th 2008, 02:09 PM
bobak

well you cant use log values form earlier.

but a is clearly 1

thanks Mr F

edit: you added the spoiler while i typed this
February 28th 2008, 02:12 PM
bobak

On the topic of spoilers the forum really should have hide/show tags.