1. ## Algebra!

Hello folks!

I'm new here, and this is my question:

If $\displaystyle x = 10^{-3}$ so:
$\displaystyle \frac{(0,1).(0,001).10^{-1} }{10.(0,0001)}$ is?

I turned all the decimal numbers on '10', and I get $\displaystyle \frac{x}{100}$, but i believe it's not the right result

2. $\displaystyle \frac{0.1 \cdot 0.001 \cdot 10^{-1}}{10 \cdot 0.0001} = \frac{10^{-1} \cdot 10^{-2} \cdot 10^{-1}}{10 \cdot 10^{-4}}$

$\displaystyle = \frac{10^{-1 - 2 - 1}}{10^{1 - 4}}$

$\displaystyle = \frac{10^{-4}}{10^{-3}}$

$\displaystyle = 10^{-4 - (-3)}$

$\displaystyle = 10^{-1}$

$\displaystyle = \left(10^{-3}\right)^{\frac{1}{3}}$

$\displaystyle = x^{\frac{1}{3}}$.

3. Thanx!

But I didn't understand why $\displaystyle 0,001=10^{-2}$... I believe it's $\displaystyle 10^{-3}$...
This will change the way to make it?

PS: I saw the answer in the end of the book () and is $\displaystyle 10x$! How can I do this????

4. Originally Posted by PxEckx
Thanx!

But I didn't understand why $\displaystyle 0,001=10^{-2}$... I believe it's $\displaystyle 10^{-3}$...
You are correct. The power of ten for a digit in a number can always be found by moving the comma all the way to the right side of the digit in question one digit at a time, counting the number of times you have to move it. An overall rightward motion describes negative powers while leftward motion describes positive powers. You have to move it over 3 times to the right to get it to the right side of the 1, so this number can be written as $\displaystyle 1\times 10^{-3} = 10^{-3}$.
We can rewrite your entire number as
$\displaystyle \frac{10^{-1}\cdot 10^{-3}\cdot 10^{-1}}{10^1\cdot 10^{-4}}$
You can now use what you know about exponents: Multiplication of two powers preserves the base and adds the exponents. similarly, division of two powers preserves the base and subtracts the power of the denominator from the power of the numerator. You could also note that division by a number is the same as multiplication by its reciprocal. Use whichever you feel most comfortable with:
$\displaystyle = \frac{10^{(-1) + (-3) + (-1)}}{10^{1 + (-4)}}$
$\displaystyle = \frac{10^{-5}}{10^{-3}}$
$\displaystyle = 10^{(-5) - (-3)} = 10^{-5 + 3} = 10^{-2}$
Since $\displaystyle x = 10^{-3}$, we have $\displaystyle 10x = 10\cdot 10^{-3} = 10^{-3 + 1} = 10^{-2}$.

5. Yes, that was a typo. Thanks.