# Thread: Log Question

1. ## Log Question

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

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

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

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

[OCR STEP I 2000, Question 1]
I could do with a tiny bit of guidance on this. thanks in advance for any help

- Bobak

2. Hello, bobak!

Here's the first one . . .

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

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

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

$\displaystyle (b)\;\;\log_{10}(6) \;=\;\log_{10}(2\cdot3) \;=\;\log_{10}(2)\;+ \;\log_{10}(3) \;=\;0.30103 \;+\;0.47712 \;\approx \;0.778$

3. 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 $\displaystyle 47 + \log_{10} 5 < 100 \log_{10} 3 < 47 + \log_{10} 6$.

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

So clearly $\displaystyle 3^{100} = 5.abcd..... \times 10^{47}$ - its first digit is now obvious.

4. 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 $\displaystyle a \times 10^x < 2^{1000} < (a + 1) \times 10^x$, where obviously a and x aren't known yet!

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

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

$\displaystyle 301 + \log_{10} a < 301.029996 < 301 + \log_{10} (a + 1)$.

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

Answer: a = 1 so the first digit is 1.

5. 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

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