Solving $2^x \cdot 5^y = 0,128$

Question

$x,y \in \mathbb Z$

$2^x \cdot 5^y = 0,128$

$x+y = ?$

My attempt:

I know that

$$0,128 = \frac{128}{1000}$$

$$5^3 = 125$$

$$2^{-3} = \frac{1}{8}$$

EDIT:

$2^7 = 128$

Then we need to get

$0,128$

$2^{-3}=\frac{1}{8}$ – cansomeonehelpmeout Mar 18 '18 at 15:33 — cansomeonehelpmeout, Mar 18 '18 at 15:33
@cansomeonehelpmeout missed it, nice catch. – Mar 18 '18 at 15:33 — , Mar 18 '18 at 15:33
Does it help if I point out that $2^7=128$? – MJD Mar 18 '18 at 15:34 — MJD, Mar 18 '18 at 15:34
@MJD Now, got it better. The thing is to get $0,128$. – Mar 18 '18 at 15:35 — , Mar 18 '18 at 15:35

score 1 · Accepted Answer · answered Mar 18 '18 at 15:28

1

Hint: $0.128=2^7\times 10^{-3}=2^4\times 5^{-3}$

answered Mar 18 '18 at 15:28

CY Aries

How? I truly didn't get what you mean. – Mar 18 '18 at 15:30
You can find $x$ and $y$ with that. – CY Aries Mar 18 '18 at 15:35
Yes, found that! However, I'd be grateful If you explain what you've done. – Mar 18 '18 at 15:37
$2^x5^y=2^45^{-3}$. So, we have $2^x5^{y+3}=2^4$. As prime decomposition of an integer is unique, $x=4$ and $y+3=0$. – CY Aries Mar 18 '18 at 15:47
Hold on, why did you use $5^{-3}$ instead of $10^{-3}$? – Mar 18 '18 at 15:48
$2^7\times 10^{-3}=2^7\times 2^{-3}\times 5^{-3}=2^4\times 5^{-3}$ – CY Aries Mar 18 '18 at 15:50
I need prime decomposition to ensure the uniqueness. So we have to use $5$ instead of $10$. – CY Aries Mar 18 '18 at 15:52

cansomeonehelpmeout · Answer 2 · 2018-03-18T15:52:17.037

0

You know that $128=2^7$, and $1000=10^3=(2\cdot 5)^3=2^3\cdot 5^3$. Therefore

$$0.128=\frac{128}{1000}=\frac{2^7}{2^3\cdot 5^3}$$

Can you do the rest?

edited Mar 18 '18 at 15:52

answered Mar 18 '18 at 15:35

For $1000$, we get $10^3$ hence $\frac{2^7}{10^3}$ and? – Mar 18 '18 at 15:38
@Enzo, good! Now if you can write $10=2^x\cdot 5^y$ then $10^3=2^{3x}\cdot 5^{3y}$. – cansomeonehelpmeout Mar 18 '18 at 15:39

score 0 · Answer 3 · answered Mar 18 '18 at 16:23

0

You can write 5 in terms of 10 and 2... so you can write the problem as: $2^x \cdot (\frac{10}{2})^y=.128$

$2^{x-y} \cdot 10^y=.128$

Find $y$ first and then $x $

answered Mar 18 '18 at 16:23

randomgirl

3 Answers3