Simplifying equations while holding some constraints
up vote
1
down vote
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I have two equation and i tried to solve them but i can't simplify them :
- $(frac{1}{b})^{frac{1}{r}} = 0.8$
- $ br = n$
$r = frac{n}{b} rightarrow frac{r}{n} = frac{1}{b}$.
Plugging this equation in first one i have:
- $(frac{r}{n})^{frac{1}{r}} = 0.8$
Now i have :
$frac{r}{n} = {(0.8)}^{r}$
Now suppose i know the value of $n = 20:$
How can i derive the equation in terms of $r$?
exponential-function systems-of-equations exponentiation
edited Nov 13 at 12:50
Harry Peter
5,43111439
asked Nov 3 at 3:50
Khan Saab
1378
add a comment |
up vote
1
down vote
favorite
I have two equation and i tried to solve them but i can't simplify them :
- $(frac{1}{b})^{frac{1}{r}} = 0.8$
- $ br = n$
$r = frac{n}{b} rightarrow frac{r}{n} = frac{1}{b}$.
Plugging this equation in first one i have:
- $(frac{r}{n})^{frac{1}{r}} = 0.8$
Now i have :
$frac{r}{n} = {(0.8)}^{r}$
Now suppose i know the value of $n = 20:$
How can i derive the equation in terms of $r$?
exponential-function systems-of-equations exponentiation
edited Nov 13 at 12:50
Harry Peter
5,43111439
asked Nov 3 at 3:50
Khan Saab
1378
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have two equation and i tried to solve them but i can't simplify them :
- $(frac{1}{b})^{frac{1}{r}} = 0.8$
- $ br = n$
$r = frac{n}{b} rightarrow frac{r}{n} = frac{1}{b}$.
Plugging this equation in first one i have:
- $(frac{r}{n})^{frac{1}{r}} = 0.8$
Now i have :
$frac{r}{n} = {(0.8)}^{r}$
Now suppose i know the value of $n = 20:$
How can i derive the equation in terms of $r$?
exponential-function systems-of-equations exponentiation
edited Nov 13 at 12:50
Harry Peter
5,43111439
asked Nov 3 at 3:50
Khan Saab
1378
I have two equation and i tried to solve them but i can't simplify them :
- $(frac{1}{b})^{frac{1}{r}} = 0.8$
- $ br = n$
$r = frac{n}{b} rightarrow frac{r}{n} = frac{1}{b}$.
Plugging this equation in first one i have:
- $(frac{r}{n})^{frac{1}{r}} = 0.8$
Now i have :
$frac{r}{n} = {(0.8)}^{r}$
Now suppose i know the value of $n = 20:$
How can i derive the equation in terms of $r$?
exponential-function systems-of-equations exponentiation
exponential-function systems-of-equations exponentiation
edited Nov 13 at 12:50
Harry Peter
5,43111439
asked Nov 3 at 3:50
Khan Saab
1378
edited Nov 13 at 12:50
Harry Peter
5,43111439
asked Nov 3 at 3:50
Khan Saab
1378
edited Nov 13 at 12:50
Harry Peter
5,43111439
edited Nov 13 at 12:50
Harry Peter
5,43111439
edited Nov 13 at 12:50
Harry Peter
5,43111439
5,43111439
asked Nov 3 at 3:50
Khan Saab
1378
asked Nov 3 at 3:50
Khan Saab
1378
asked Nov 3 at 3:50
Khan Saab
1378
1378
add a comment |
add a comment |
1 Answer
1
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oldest
votes
up vote
1
down vote
You will need the Lambert W function for this, which is non-elementary ($W(xe^x)=x$):
$$r/n=0.8^r=e^{rln0.8}$$
$$re^{-rln0.8}=n$$
$$re^{rln1.25}=n$$
$$(rln1.25)e^{rln1.25}=nln1.25$$
$$rln1.25=W(nln1.25)$$
$$r=frac{W(nln1.25)}{ln1.25}$$
If $n=20$ then $r=5.658289dots$
answered Nov 3 at 4:04
Parcly Taxel
41k137199
Thank you very much. I never heard of this method even i had many advanced mathematical courses :(
– Khan Saab
Nov 3 at 4:06
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You will need the Lambert W function for this, which is non-elementary ($W(xe^x)=x$):
$$r/n=0.8^r=e^{rln0.8}$$
$$re^{-rln0.8}=n$$
$$re^{rln1.25}=n$$
$$(rln1.25)e^{rln1.25}=nln1.25$$
$$rln1.25=W(nln1.25)$$
$$r=frac{W(nln1.25)}{ln1.25}$$
If $n=20$ then $r=5.658289dots$
answered Nov 3 at 4:04
Parcly Taxel
41k137199
Thank you very much. I never heard of this method even i had many advanced mathematical courses :(
– Khan Saab
Nov 3 at 4:06
add a comment |
up vote
1
down vote
You will need the Lambert W function for this, which is non-elementary ($W(xe^x)=x$):
$$r/n=0.8^r=e^{rln0.8}$$
$$re^{-rln0.8}=n$$
$$re^{rln1.25}=n$$
$$(rln1.25)e^{rln1.25}=nln1.25$$
$$rln1.25=W(nln1.25)$$
$$r=frac{W(nln1.25)}{ln1.25}$$
If $n=20$ then $r=5.658289dots$
answered Nov 3 at 4:04
Parcly Taxel
41k137199
Thank you very much. I never heard of this method even i had many advanced mathematical courses :(
– Khan Saab
Nov 3 at 4:06
add a comment |
up vote
1
down vote
up vote
1
down vote
You will need the Lambert W function for this, which is non-elementary ($W(xe^x)=x$):
$$r/n=0.8^r=e^{rln0.8}$$
$$re^{-rln0.8}=n$$
$$re^{rln1.25}=n$$
$$(rln1.25)e^{rln1.25}=nln1.25$$
$$rln1.25=W(nln1.25)$$
$$r=frac{W(nln1.25)}{ln1.25}$$
If $n=20$ then $r=5.658289dots$
answered Nov 3 at 4:04
Parcly Taxel
41k137199
You will need the Lambert W function for this, which is non-elementary ($W(xe^x)=x$):
$$r/n=0.8^r=e^{rln0.8}$$
$$re^{-rln0.8}=n$$
$$re^{rln1.25}=n$$
$$(rln1.25)e^{rln1.25}=nln1.25$$
$$rln1.25=W(nln1.25)$$
$$r=frac{W(nln1.25)}{ln1.25}$$
If $n=20$ then $r=5.658289dots$
answered Nov 3 at 4:04
Parcly Taxel
41k137199
answered Nov 3 at 4:04
Parcly Taxel
41k137199
answered Nov 3 at 4:04
Parcly Taxel
41k137199
answered Nov 3 at 4:04
Parcly Taxel
41k137199
41k137199
Thank you very much. I never heard of this method even i had many advanced mathematical courses :(
– Khan Saab
Nov 3 at 4:06
add a comment |
Thank you very much. I never heard of this method even i had many advanced mathematical courses :(
– Khan Saab
Nov 3 at 4:06
Thank you very much. I never heard of this method even i had many advanced mathematical courses :(
– Khan Saab
Nov 3 at 4:06
Thank you very much. I never heard of this method even i had many advanced mathematical courses :(
– Khan Saab
Nov 3 at 4:06
add a comment |
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