# Operations for operands with powers of different bases

How can I quickly simplify or resolve with no calculator an operation that involves powers with different bases? For example:

$$(2^{128})/(2\cdot10^{19})$$

I thought of converting $2\cdot10^{19}$ to a power base two. Something like $2\cdot(3.16^2)^{19}$ My intent was to have the same base so I could subtract the exponents:

$$128-(2\cdot19)\text{ or }128-38$$

But I don't think I'm on the right track.

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Don't write $2^{128}/2\cdot10^{19}$ if you mean $2^{128}/(2\cdot10^{19})$. In standard usage, $2^{128}/2\cdot10^{19}$ means $(2^{128}/2)\cdot10^{19}$. – Michael Hardy Dec 5 '12 at 15:01
If you want just an estimate on this, you can approximate $2^{10} = 1024 \approx 10^3$, hence $10^{19} \approx 2^{60} \cdot 10$. – martini Dec 5 '12 at 15:07

$$\frac{2^{128}}{2\cdot10^{19}}=\frac{2^{128}}{2\cdot2^{19}\cdot5^{19}}=\frac{2^{108}}{5^{19}}$$

This can't be simplified any further as $2$ and $5$ are prime numbers.

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Thanks, I know this is not precise at all, but is it valid to approximate $5^{19}$ to base 2 by doing $(2.23^2)^{19}$ – Ulises Dec 5 '12 at 15:04
Not only is $5\neq 2.23^2$, there is also no point in writing it that way. In general, the most reduced form of writing rational numbers is in terms of powers of primes. – Sarastro Dec 5 '12 at 15:10
ok, thanks for your answer. The only reason why I wanted to do it is to have an approximate power of 2 (just compare with other stuff in my "computer science" class) – Ulises Dec 5 '12 at 16:58

If the bases in a quotient or product are equivalent, you can manipulate exponents (subtracting for in the case of quotients, adding in the case of products). If two distinct bases have equivalent exponents, you can manipulate the bases: for $\dfrac{a^7}{b^7} = \left(\dfrac ab\right)^7$ and $a^7\cdot b^7 = (a\cdot b)^7$

$$\frac{2^{128}}{2\cdot10^{19}} = \frac{2^{(128-1)}}{10^{19}}= \frac{2^{127}}{10^{19}} = \frac{2^{127}}{(2 \cdot 5)^{19}}= \frac{2^{127}}{2^{19}\cdot 5^{19}} = \frac{2^{(127 - 19)}}{5^{19}} = \frac{2^{108}}{5^{19}}$$
That's as simpified as it gets, without computing $2^{108}$ and $5^{19}$, since $2$ and $5$ are both prime (hence coprime!).
If you meant to write: $\dfrac{2^{128}}{(2\cdot10)^{19}}$, then: $$\frac{2^{128}}{(2\cdot10)^{19}} = \frac{2^{128}}{2^{19}\cdot 10^{19}} = \frac{2^{128}}{2^{19}\cdot 2^{19} \cdot 5^{19}} = \frac{2^{128- 19 - 19}}{5^{19}} = \frac{2^{90}}{5^{19}}$$
As you see, there is a very big difference between: $\quad \dfrac{2^{128}}{(2\cdot10)^{19}}\quad$ and $\quad\dfrac{2^{128}}{2\cdot10^{19}}$
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