Finding the inverse of my f(x) function that includes a fractional exponent (with a variable in the fraction)
$begingroup$
Im a noob mathematician.
I am trying to find the inverse of this function:
$$F(x) = frac{x + 299 cdot 2^frac{x-1}{7}}{4} $$
In the same way that: $$f(y) = ½(y - 1)$$ is the inverse of: $$f(x) = 2x + 1$$
I know the procedure goes along like this:
1) Let y = f(x)
2) Swap the x"s and y"s
3) Solve with y as the subject
The problem im having is with the exponent: (x-1)/7. I'm not sure what to do with it so that I can solve for y.
If someone could solve this for me that would be great. I am also interested in the steps to the solution, so including the steps would be a bonus.
Thanks in advance!
logarithms exponentiation inverse-function
asked Dec 29 '18 at 4:13
Houssein KHoussein K
111
$endgroup$
add a comment |
$begingroup$
Im a noob mathematician.
I am trying to find the inverse of this function:
$$F(x) = frac{x + 299 cdot 2^frac{x-1}{7}}{4} $$
In the same way that: $$f(y) = ½(y - 1)$$ is the inverse of: $$f(x) = 2x + 1$$
I know the procedure goes along like this:
1) Let y = f(x)
2) Swap the x"s and y"s
3) Solve with y as the subject
The problem im having is with the exponent: (x-1)/7. I'm not sure what to do with it so that I can solve for y.
If someone could solve this for me that would be great. I am also interested in the steps to the solution, so including the steps would be a bonus.
Thanks in advance!
logarithms exponentiation inverse-function
asked Dec 29 '18 at 4:13
Houssein KHoussein K
111
$endgroup$
$begingroup$
The Lambert $W$ function, also known as the productlog function, might be worth looking at. reference.wolfram.com/language/ref/ProductLog.html & en.wikipedia.org/wiki/Lambert_W_function
$endgroup$
– Eevee Trainer
Dec 29 '18 at 4:18
$begingroup$
I say this because I did have Wolfram Alpha calculate the inverse (goo.gl/Qy4ehs) and it uses the $W$ function. So finding this inverse is not going to be trivial. I'm trying to figure out how to do so myself but the $W$ function isn't my area of expertise. (Deleted and reposted this comment because I found the inverse of the wrong thing and then had trouble linking to it. Still uses the $W$ function though.)
$endgroup$
– Eevee Trainer
Dec 29 '18 at 4:24
1
$begingroup$
@EeveeTrainer Thanks for linking to wolfram and mentioning the W function. I saw this get mentioned in other inverse-function questions, but since they did not exactly match my exponent I was hoping it wouldn't apply to me. sigh I guess I have to learn something new now.
$endgroup$
– Houssein K
Dec 29 '18 at 4:29
$begingroup$
Well learning something new is always a good thing.
$endgroup$
– Mohammad Zuhair Khan
Dec 29 '18 at 4:42
$begingroup$
The substitution $u=x+4F(x)$ greatly simplifies the equation.
$endgroup$
– Ahmed S. Attaalla
Dec 29 '18 at 6:25
add a comment |
$begingroup$
Im a noob mathematician.
I am trying to find the inverse of this function:
$$F(x) = frac{x + 299 cdot 2^frac{x-1}{7}}{4} $$
In the same way that: $$f(y) = ½(y - 1)$$ is the inverse of: $$f(x) = 2x + 1$$
I know the procedure goes along like this:
1) Let y = f(x)
2) Swap the x"s and y"s
3) Solve with y as the subject
The problem im having is with the exponent: (x-1)/7. I'm not sure what to do with it so that I can solve for y.
If someone could solve this for me that would be great. I am also interested in the steps to the solution, so including the steps would be a bonus.
Thanks in advance!
logarithms exponentiation inverse-function
asked Dec 29 '18 at 4:13
Houssein KHoussein K
111
$endgroup$
Im a noob mathematician.
I am trying to find the inverse of this function:
$$F(x) = frac{x + 299 cdot 2^frac{x-1}{7}}{4} $$
In the same way that: $$f(y) = ½(y - 1)$$ is the inverse of: $$f(x) = 2x + 1$$
I know the procedure goes along like this:
1) Let y = f(x)
2) Swap the x"s and y"s
3) Solve with y as the subject
The problem im having is with the exponent: (x-1)/7. I'm not sure what to do with it so that I can solve for y.
If someone could solve this for me that would be great. I am also interested in the steps to the solution, so including the steps would be a bonus.
Thanks in advance!
logarithms exponentiation inverse-function
logarithms exponentiation inverse-function
asked Dec 29 '18 at 4:13
Houssein KHoussein K
111
asked Dec 29 '18 at 4:13
Houssein KHoussein K
111
asked Dec 29 '18 at 4:13
Houssein KHoussein K
111
asked Dec 29 '18 at 4:13
Houssein KHoussein K
111
asked Dec 29 '18 at 4:13
Houssein KHoussein K
111
111
$begingroup$
The Lambert $W$ function, also known as the productlog function, might be worth looking at. reference.wolfram.com/language/ref/ProductLog.html & en.wikipedia.org/wiki/Lambert_W_function
$endgroup$
– Eevee Trainer
Dec 29 '18 at 4:18
$begingroup$
I say this because I did have Wolfram Alpha calculate the inverse (goo.gl/Qy4ehs) and it uses the $W$ function. So finding this inverse is not going to be trivial. I'm trying to figure out how to do so myself but the $W$ function isn't my area of expertise. (Deleted and reposted this comment because I found the inverse of the wrong thing and then had trouble linking to it. Still uses the $W$ function though.)
$endgroup$
– Eevee Trainer
Dec 29 '18 at 4:24
1
$begingroup$
@EeveeTrainer Thanks for linking to wolfram and mentioning the W function. I saw this get mentioned in other inverse-function questions, but since they did not exactly match my exponent I was hoping it wouldn't apply to me. sigh I guess I have to learn something new now.
$endgroup$
– Houssein K
Dec 29 '18 at 4:29
$begingroup$
Well learning something new is always a good thing.
$endgroup$
– Mohammad Zuhair Khan
Dec 29 '18 at 4:42
$begingroup$
The substitution $u=x+4F(x)$ greatly simplifies the equation.
$endgroup$
– Ahmed S. Attaalla
Dec 29 '18 at 6:25
add a comment |
$begingroup$
The Lambert $W$ function, also known as the productlog function, might be worth looking at. reference.wolfram.com/language/ref/ProductLog.html & en.wikipedia.org/wiki/Lambert_W_function
$endgroup$
– Eevee Trainer
Dec 29 '18 at 4:18
$begingroup$
I say this because I did have Wolfram Alpha calculate the inverse (goo.gl/Qy4ehs) and it uses the $W$ function. So finding this inverse is not going to be trivial. I'm trying to figure out how to do so myself but the $W$ function isn't my area of expertise. (Deleted and reposted this comment because I found the inverse of the wrong thing and then had trouble linking to it. Still uses the $W$ function though.)
$endgroup$
– Eevee Trainer
Dec 29 '18 at 4:24
1
$begingroup$
@EeveeTrainer Thanks for linking to wolfram and mentioning the W function. I saw this get mentioned in other inverse-function questions, but since they did not exactly match my exponent I was hoping it wouldn't apply to me. sigh I guess I have to learn something new now.
$endgroup$
– Houssein K
Dec 29 '18 at 4:29
$begingroup$
Well learning something new is always a good thing.
$endgroup$
– Mohammad Zuhair Khan
Dec 29 '18 at 4:42
$begingroup$
The substitution $u=x+4F(x)$ greatly simplifies the equation.
$endgroup$
– Ahmed S. Attaalla
Dec 29 '18 at 6:25
$begingroup$
The Lambert $W$ function, also known as the productlog function, might be worth looking at. reference.wolfram.com/language/ref/ProductLog.html & en.wikipedia.org/wiki/Lambert_W_function
$endgroup$
– Eevee Trainer
Dec 29 '18 at 4:18
$begingroup$
The Lambert $W$ function, also known as the productlog function, might be worth looking at. reference.wolfram.com/language/ref/ProductLog.html & en.wikipedia.org/wiki/Lambert_W_function
$endgroup$
– Eevee Trainer
Dec 29 '18 at 4:18
$begingroup$
I say this because I did have Wolfram Alpha calculate the inverse (goo.gl/Qy4ehs) and it uses the $W$ function. So finding this inverse is not going to be trivial. I'm trying to figure out how to do so myself but the $W$ function isn't my area of expertise. (Deleted and reposted this comment because I found the inverse of the wrong thing and then had trouble linking to it. Still uses the $W$ function though.)
$endgroup$
– Eevee Trainer
Dec 29 '18 at 4:24
$begingroup$
I say this because I did have Wolfram Alpha calculate the inverse (goo.gl/Qy4ehs) and it uses the $W$ function. So finding this inverse is not going to be trivial. I'm trying to figure out how to do so myself but the $W$ function isn't my area of expertise. (Deleted and reposted this comment because I found the inverse of the wrong thing and then had trouble linking to it. Still uses the $W$ function though.)
$endgroup$
– Eevee Trainer
Dec 29 '18 at 4:24
1
1
$begingroup$
@EeveeTrainer Thanks for linking to wolfram and mentioning the W function. I saw this get mentioned in other inverse-function questions, but since they did not exactly match my exponent I was hoping it wouldn't apply to me. sigh I guess I have to learn something new now.
$endgroup$
– Houssein K
Dec 29 '18 at 4:29
$begingroup$
@EeveeTrainer Thanks for linking to wolfram and mentioning the W function. I saw this get mentioned in other inverse-function questions, but since they did not exactly match my exponent I was hoping it wouldn't apply to me. sigh I guess I have to learn something new now.
$endgroup$
– Houssein K
Dec 29 '18 at 4:29
$begingroup$
Well learning something new is always a good thing.
$endgroup$
– Mohammad Zuhair Khan
Dec 29 '18 at 4:42
$begingroup$
Well learning something new is always a good thing.
$endgroup$
– Mohammad Zuhair Khan
Dec 29 '18 at 4:42
$begingroup$
The substitution $u=x+4F(x)$ greatly simplifies the equation.
$endgroup$
– Ahmed S. Attaalla
Dec 29 '18 at 6:25
$begingroup$
The substitution $u=x+4F(x)$ greatly simplifies the equation.
$endgroup$
– Ahmed S. Attaalla
Dec 29 '18 at 6:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In general the equation,
$$e^{-cx}=a_0left(x-r right)$$
Has solution,
$$r+frac{1}{c} W(frac{ce^{-cr}}{a_0})=x$$
The reason is below this line.
Using the lambert W function. Let $x-r=u$ so that $x=r+u$.
$$e^{-c(r+u)}=a_0 u $$
$$e^{-cr}e^{-cu}=a_0u$$
$$e^{-cu}=a_0e^{cr} u$$
$$1=a_0e^{cr} u e^{cu}$$
$$frac{e^{-cr}}{a_0}=u e^{cu}$$
$$frac{ce^{-cr}}{a_0}=cu e^{cu}$$
$$W(frac{ce^{-cr}}{a_0})=cu$$
$$frac{1}{c} W(frac{ce^{-cr}}{a_0})=u$$
$$r+frac{1}{c} W(frac{ce^{-cr}}{a_0})=x$$
The equation you have written can be rearranged to the form considered previously.
$$4F=x+299(2^{-frac{1}{7}}) e^{x frac{ln 2}{7}}$$
$$frac{4F-x}{299}=2^{-frac{1}{7}} e^{x frac{ln 2}{7}}$$
$$frac{2^{frac{1}{7}}}{299} left( 4F-x right)=e^{ x frac{ln 2}{7}}$$
$$frac{-2^{frac{1}{7}}}{299} left( x-4F right)=e^{ x frac{ln 2}{7}}$$
Thus, to find the inverse use $r=4x$, $a_0= frac{-2^{frac{1}{7}}}{299}$, and $c=-frac{ln 2}{7}$.
answered Dec 29 '18 at 6:00
Ahmed S. AttaallaAhmed S. Attaalla
14.8k12150
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
In general the equation,
$$e^{-cx}=a_0left(x-r right)$$
Has solution,
$$r+frac{1}{c} W(frac{ce^{-cr}}{a_0})=x$$
The reason is below this line.
Using the lambert W function. Let $x-r=u$ so that $x=r+u$.
$$e^{-c(r+u)}=a_0 u $$
$$e^{-cr}e^{-cu}=a_0u$$
$$e^{-cu}=a_0e^{cr} u$$
$$1=a_0e^{cr} u e^{cu}$$
$$frac{e^{-cr}}{a_0}=u e^{cu}$$
$$frac{ce^{-cr}}{a_0}=cu e^{cu}$$
$$W(frac{ce^{-cr}}{a_0})=cu$$
$$frac{1}{c} W(frac{ce^{-cr}}{a_0})=u$$
$$r+frac{1}{c} W(frac{ce^{-cr}}{a_0})=x$$
The equation you have written can be rearranged to the form considered previously.
$$4F=x+299(2^{-frac{1}{7}}) e^{x frac{ln 2}{7}}$$
$$frac{4F-x}{299}=2^{-frac{1}{7}} e^{x frac{ln 2}{7}}$$
$$frac{2^{frac{1}{7}}}{299} left( 4F-x right)=e^{ x frac{ln 2}{7}}$$
$$frac{-2^{frac{1}{7}}}{299} left( x-4F right)=e^{ x frac{ln 2}{7}}$$
Thus, to find the inverse use $r=4x$, $a_0= frac{-2^{frac{1}{7}}}{299}$, and $c=-frac{ln 2}{7}$.
answered Dec 29 '18 at 6:00
Ahmed S. AttaallaAhmed S. Attaalla
14.8k12150
$endgroup$
add a comment |
$begingroup$
In general the equation,
$$e^{-cx}=a_0left(x-r right)$$
Has solution,
$$r+frac{1}{c} W(frac{ce^{-cr}}{a_0})=x$$
The reason is below this line.
Using the lambert W function. Let $x-r=u$ so that $x=r+u$.
$$e^{-c(r+u)}=a_0 u $$
$$e^{-cr}e^{-cu}=a_0u$$
$$e^{-cu}=a_0e^{cr} u$$
$$1=a_0e^{cr} u e^{cu}$$
$$frac{e^{-cr}}{a_0}=u e^{cu}$$
$$frac{ce^{-cr}}{a_0}=cu e^{cu}$$
$$W(frac{ce^{-cr}}{a_0})=cu$$
$$frac{1}{c} W(frac{ce^{-cr}}{a_0})=u$$
$$r+frac{1}{c} W(frac{ce^{-cr}}{a_0})=x$$
The equation you have written can be rearranged to the form considered previously.
$$4F=x+299(2^{-frac{1}{7}}) e^{x frac{ln 2}{7}}$$
$$frac{4F-x}{299}=2^{-frac{1}{7}} e^{x frac{ln 2}{7}}$$
$$frac{2^{frac{1}{7}}}{299} left( 4F-x right)=e^{ x frac{ln 2}{7}}$$
$$frac{-2^{frac{1}{7}}}{299} left( x-4F right)=e^{ x frac{ln 2}{7}}$$
Thus, to find the inverse use $r=4x$, $a_0= frac{-2^{frac{1}{7}}}{299}$, and $c=-frac{ln 2}{7}$.
answered Dec 29 '18 at 6:00
Ahmed S. AttaallaAhmed S. Attaalla
14.8k12150
$endgroup$
add a comment |
$begingroup$
In general the equation,
$$e^{-cx}=a_0left(x-r right)$$
Has solution,
$$r+frac{1}{c} W(frac{ce^{-cr}}{a_0})=x$$
The reason is below this line.
Using the lambert W function. Let $x-r=u$ so that $x=r+u$.
$$e^{-c(r+u)}=a_0 u $$
$$e^{-cr}e^{-cu}=a_0u$$
$$e^{-cu}=a_0e^{cr} u$$
$$1=a_0e^{cr} u e^{cu}$$
$$frac{e^{-cr}}{a_0}=u e^{cu}$$
$$frac{ce^{-cr}}{a_0}=cu e^{cu}$$
$$W(frac{ce^{-cr}}{a_0})=cu$$
$$frac{1}{c} W(frac{ce^{-cr}}{a_0})=u$$
$$r+frac{1}{c} W(frac{ce^{-cr}}{a_0})=x$$
The equation you have written can be rearranged to the form considered previously.
$$4F=x+299(2^{-frac{1}{7}}) e^{x frac{ln 2}{7}}$$
$$frac{4F-x}{299}=2^{-frac{1}{7}} e^{x frac{ln 2}{7}}$$
$$frac{2^{frac{1}{7}}}{299} left( 4F-x right)=e^{ x frac{ln 2}{7}}$$
$$frac{-2^{frac{1}{7}}}{299} left( x-4F right)=e^{ x frac{ln 2}{7}}$$
Thus, to find the inverse use $r=4x$, $a_0= frac{-2^{frac{1}{7}}}{299}$, and $c=-frac{ln 2}{7}$.
answered Dec 29 '18 at 6:00
Ahmed S. AttaallaAhmed S. Attaalla
14.8k12150
$endgroup$
In general the equation,
$$e^{-cx}=a_0left(x-r right)$$
Has solution,
$$r+frac{1}{c} W(frac{ce^{-cr}}{a_0})=x$$
The reason is below this line.
Using the lambert W function. Let $x-r=u$ so that $x=r+u$.
$$e^{-c(r+u)}=a_0 u $$
$$e^{-cr}e^{-cu}=a_0u$$
$$e^{-cu}=a_0e^{cr} u$$
$$1=a_0e^{cr} u e^{cu}$$
$$frac{e^{-cr}}{a_0}=u e^{cu}$$
$$frac{ce^{-cr}}{a_0}=cu e^{cu}$$
$$W(frac{ce^{-cr}}{a_0})=cu$$
$$frac{1}{c} W(frac{ce^{-cr}}{a_0})=u$$
$$r+frac{1}{c} W(frac{ce^{-cr}}{a_0})=x$$
The equation you have written can be rearranged to the form considered previously.
$$4F=x+299(2^{-frac{1}{7}}) e^{x frac{ln 2}{7}}$$
$$frac{4F-x}{299}=2^{-frac{1}{7}} e^{x frac{ln 2}{7}}$$
$$frac{2^{frac{1}{7}}}{299} left( 4F-x right)=e^{ x frac{ln 2}{7}}$$
$$frac{-2^{frac{1}{7}}}{299} left( x-4F right)=e^{ x frac{ln 2}{7}}$$
Thus, to find the inverse use $r=4x$, $a_0= frac{-2^{frac{1}{7}}}{299}$, and $c=-frac{ln 2}{7}$.
answered Dec 29 '18 at 6:00
Ahmed S. AttaallaAhmed S. Attaalla
14.8k12150
edited Dec 29 '18 at 6:36
answered Dec 29 '18 at 6:00
Ahmed S. AttaallaAhmed S. Attaalla
14.8k12150
answered Dec 29 '18 at 6:00
Ahmed S. AttaallaAhmed S. Attaalla
14.8k12150
answered Dec 29 '18 at 6:00
Ahmed S. AttaallaAhmed S. Attaalla
14.8k12150
14.8k12150
add a comment |
add a comment |
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$begingroup$
The Lambert $W$ function, also known as the productlog function, might be worth looking at. reference.wolfram.com/language/ref/ProductLog.html & en.wikipedia.org/wiki/Lambert_W_function
$endgroup$
– Eevee Trainer
Dec 29 '18 at 4:18
$begingroup$
I say this because I did have Wolfram Alpha calculate the inverse (goo.gl/Qy4ehs) and it uses the $W$ function. So finding this inverse is not going to be trivial. I'm trying to figure out how to do so myself but the $W$ function isn't my area of expertise. (Deleted and reposted this comment because I found the inverse of the wrong thing and then had trouble linking to it. Still uses the $W$ function though.)
$endgroup$
– Eevee Trainer
Dec 29 '18 at 4:24
1
$begingroup$
@EeveeTrainer Thanks for linking to wolfram and mentioning the W function. I saw this get mentioned in other inverse-function questions, but since they did not exactly match my exponent I was hoping it wouldn't apply to me. sigh I guess I have to learn something new now.
$endgroup$
– Houssein K
Dec 29 '18 at 4:29
$begingroup$
Well learning something new is always a good thing.
$endgroup$
– Mohammad Zuhair Khan
Dec 29 '18 at 4:42
$begingroup$
The substitution $u=x+4F(x)$ greatly simplifies the equation.
$endgroup$
– Ahmed S. Attaalla
Dec 29 '18 at 6:25