Logarithm: octave and decade I am having some difficulty of finding a common divisor between an octave and a decade.
The doubling of a value say from $1$ to $2$ corresponds to a difference of $20\, \log_{10}(2) = 6.020600$ dB to $6$ d.p. So there is one Octave difference between these two amplitude values. Let us divide this dB value into three. This corresponds to dividing the interval between $1$ and $2$ into three equal intervals. One such interval will represent then $2.006867$ dB. This corresponds to a ratio of $10^{\frac{2.006867}{20}}=1.259921.$
So if I take the number $1$ and multiply by $1.259921$ three times in succession, I get the value of $2$ (actually $2.000000$ when performing the calculation).
I note that $2^{\frac{1}{3}}$ is equal to the ratio $1.259921$ obtained above to $6$ d.p. 
I now move on to the Decade. If I consider the numbers $1$ and $10$, there is a difference of $20\, \log_{10}(10) = 20$ dB. There is one Decade difference between the two amplitude values $1$ and $10$. I divide this interval into $10$ equal parts with each part thus corresponding to $2$ dB. This corresponds to a ratio of $10^{\frac{2}{20}}=1.258925$ to $6$ d.p. So taking the number $1$ and multiplying it by $1.258925$ ten times successively yields the value $10$ (in fact $9.99967$).
From the above results for the Octave part and Decade part respectively, we observe that:
$$2^{\frac{1}{3}}\simeq 1.259921 \neq 10^{\frac{2}{20}} \quad \text{i.e.} \quad 1.258925$$
Had I used $2^{\frac{1}{3}}\simeq 1.259921$ in my discussion for the Decade, then multiplying the number $1$ by $2^{\frac{1}{3}}$ in succession for ten times would produce $8 \times 2^{\frac{1}{3}} = 10.079368$ and not exactly $10$.
My question is as follows:
When I divide a distance representing an Octave into $3$ intervals, a multiplication factor at each interval of $2^{\frac{1}{3}}$ gives exactly one Octave after the $3$ intervals (i.e. the number $1$ becoming $2$) but why this multiplication factor of $2^{\frac{1}{3}}$ cannot produce an accurate result for a distance of one Decade given that a decade is a $10$ times ratio? We have seen that using $2^{\frac{1}{3}}$ as the multiplying factor will take a number $1$ and ends with $10.079368$. If I am dividing an Octave into $3$ parts, then a Decade must be $10$ parts. The ratio of multiplication factor of $2^{\frac{1}{3}}$ works perfectly for an Octave but strangely not for a Decade.
 A: Remember $3dB$ switches for volume controls? The absolute minimum power difference the ear can detect is about $\phi=1.618...$ and $2$ is barely above that as far as our ability to hear the difference. Well, $10^{0.3010}=2$ and this roughly $\frac{3}{10}$ of a power of $10$ is where we get the $3$ in $3dB$. Every tactile click in a $3dB$ switch increases or decreases the sound power by a factor of $2$.
So if $10^{0.3010...}=2$, then cubing it would make $10^{0.9030...}=8$.
Note that $10^{0.1}\approx 1.259...$ so cubing it would make $10^{0.3}\approx 2$.
Therefore, taking $8*2^\frac{1}{3} $ is almost the same as $8*1.25=10$.
If you take these all together, you can see that [deci]Bells, Octaves and Decades are all just base $10$ logs unless you want to vary with base $2$ or other.
All of these logs are approximate so don't be discouraged if the don't come out exact. For a further study of logs and Bells, look here.
A: No natural power of $2$ is a power of $10$. Hence the ratio
$$\frac{\log10}{\log2}=3.32192809\cdots$$ is irrational, and not equal to $\frac{10}3$ (for the same reason that $2^{10}=1024\ne1000=10^3$ and $\sqrt[3]2\ne\sqrt[10]10$).
