# Thread: Elementary theory of logarithms

1. ## Elementary theory of logarithms

Hi,
I am wondering if anyone can help me with this exercise:
Given that $\displaystyle log2=0.3$
show that $\displaystyle log5=0.7$
I tried the following approach:
$\displaystyle 0.7=0.3+0.3+\frac{1}{3}\times0.3 \\ =\log2+\log2+\log2^\frac{1}{3}$
$\displaystyle =\log(4\times\sqrt[3]{2})$
$\displaystyle \approx\log5$

Thanks for help

Gibo

2. Originally Posted by Gibo
Hi,
I am wondering if anyone can help me with this exercise:
Given that $\displaystyle log2=0.3$
show that $\displaystyle log5=0.7$
I tried the following approach:
$\displaystyle 0.7=0.3+0.3+\frac{1}{3}\times0.3 \\ =\log2+\log2+\log2^\frac{1}{3}$
$\displaystyle =\log(4\times\sqrt[3]{2})$
$\displaystyle \approx\log5$

Thanks for help

Gibo
That mostly seems fine, but that last step seems underdeveloped. What I would do is:
$\displaystyle 0.7 = 0.3 + 0.3 + (\frac{1}{3}\*0.3) = Log2 + Log 2 + \frac{1}{3}\ Log2 = 2Log2 + \frac{1}{3}\ Log2 = Log2^2 + Log2^\frac{1}{3}\ 0.7= Log(2^2*2^\frac{1}{3}\ ) = Log2^\frac{7}{3} \approx\log 5$

Dunno, just seems a little more "fleshed-out" if you will, to me.

3. Hello, Gibo!

$\displaystyle \text{Given that }\log2\,=\,0.3,\,\text{ show that: }\log5\,=\,0.7$

Assuming that the logs are base-10 . . .

$\displaystyle \log(5) \;=\;\log\left(\dfrac{10}{2}\right) \;=\;\log(10) - \log(2) \;=\;1 - 0.3 \;=\;0.7$

4. Originally Posted by Soroban
Hello, Gibo!

Assuming that the logs are base-10 . . .

$\displaystyle \log(5) \;=\;\log\left(\dfrac{10}{2}\right) \;=\;\log(10) - \log(2) \;=\;1 - 0.3 \;=\;0.7$
...-facepalm- How did I not think of that?!

Regards,
Gibo

6. Thanks for yor help.
Regards,
Gibo