How to solve for $x$ in an equation such as “$2x^2 - 5x + 64log(x) + 776 = 0$”? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to solve equations of this form: $x^x = n$?Solving log equationHow to solve $5000 n log(n) leq 2^n/2$Solve trigonometric equation for xHow to Solve $2.3856 + log r = log(364r - 363)$solve $log m = log n + 3/2 log (1 + v/m^2)$ for mHow to solve a nonlinear equation?Solving for different variablesSolve an equation for $a$Solving the equation $2^x+2^-x = 10$
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How to solve for $x$ in an equation such as “$2x^2 - 5x + 64log(x) + 776 = 0$”?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to solve equations of this form: $x^x = n$?Solving log equationHow to solve $5000 n log(n) leq 2^n/2$Solve trigonometric equation for xHow to Solve $2.3856 + log r = log(364r - 363)$solve $log m = log n + 3/2 log (1 + v/m^2)$ for mHow to solve a nonlinear equation?Solving for different variablesSolve an equation for $a$Solving the equation $2^x+2^-x = 10$
$begingroup$
I'm trying to find the shortest distance between a point and a logarithmic line. So far I've applied the distance formula, found the derivative of the distance equation and made it equal to zero. That's how I got the equation in the title: $$2x^2 - 5x + 64log(x) + 776 = 0.$$ From this point I haven't been able to figure out how to solve for $x$. So the question is, how to I solve for $x$ in this equation?
algebra-precalculus
$endgroup$
|
show 2 more comments
$begingroup$
I'm trying to find the shortest distance between a point and a logarithmic line. So far I've applied the distance formula, found the derivative of the distance equation and made it equal to zero. That's how I got the equation in the title: $$2x^2 - 5x + 64log(x) + 776 = 0.$$ From this point I haven't been able to figure out how to solve for $x$. So the question is, how to I solve for $x$ in this equation?
algebra-precalculus
$endgroup$
1
$begingroup$
I’m pretty sure that many of these types of equations can not be solved algebraically. This isn’t to say that your solution does not have one, just to say it is not guaranteed to have one.
$endgroup$
– H Huang
Mar 26 at 16:29
$begingroup$
Alright, thanks for the answer.
$endgroup$
– Steinar Sindri Agnarsson
Mar 26 at 16:31
$begingroup$
I am writing an answer that suggests how we can analytically find a root. Should I post it?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 26 at 16:32
$begingroup$
You will need a numerical method, e.g. Newton -Raphson method
$endgroup$
– Dr. Sonnhard Graubner
Mar 26 at 16:35
$begingroup$
We find $$xapprox 5.42225066756636430174*10^-6$$
$endgroup$
– Dr. Sonnhard Graubner
Mar 26 at 16:36
|
show 2 more comments
$begingroup$
I'm trying to find the shortest distance between a point and a logarithmic line. So far I've applied the distance formula, found the derivative of the distance equation and made it equal to zero. That's how I got the equation in the title: $$2x^2 - 5x + 64log(x) + 776 = 0.$$ From this point I haven't been able to figure out how to solve for $x$. So the question is, how to I solve for $x$ in this equation?
algebra-precalculus
$endgroup$
I'm trying to find the shortest distance between a point and a logarithmic line. So far I've applied the distance formula, found the derivative of the distance equation and made it equal to zero. That's how I got the equation in the title: $$2x^2 - 5x + 64log(x) + 776 = 0.$$ From this point I haven't been able to figure out how to solve for $x$. So the question is, how to I solve for $x$ in this equation?
algebra-precalculus
algebra-precalculus
edited Mar 26 at 16:29
Steinar Sindri Agnarsson
asked Mar 26 at 16:23
Steinar Sindri AgnarssonSteinar Sindri Agnarsson
84
84
1
$begingroup$
I’m pretty sure that many of these types of equations can not be solved algebraically. This isn’t to say that your solution does not have one, just to say it is not guaranteed to have one.
$endgroup$
– H Huang
Mar 26 at 16:29
$begingroup$
Alright, thanks for the answer.
$endgroup$
– Steinar Sindri Agnarsson
Mar 26 at 16:31
$begingroup$
I am writing an answer that suggests how we can analytically find a root. Should I post it?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 26 at 16:32
$begingroup$
You will need a numerical method, e.g. Newton -Raphson method
$endgroup$
– Dr. Sonnhard Graubner
Mar 26 at 16:35
$begingroup$
We find $$xapprox 5.42225066756636430174*10^-6$$
$endgroup$
– Dr. Sonnhard Graubner
Mar 26 at 16:36
|
show 2 more comments
1
$begingroup$
I’m pretty sure that many of these types of equations can not be solved algebraically. This isn’t to say that your solution does not have one, just to say it is not guaranteed to have one.
$endgroup$
– H Huang
Mar 26 at 16:29
$begingroup$
Alright, thanks for the answer.
$endgroup$
– Steinar Sindri Agnarsson
Mar 26 at 16:31
$begingroup$
I am writing an answer that suggests how we can analytically find a root. Should I post it?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 26 at 16:32
$begingroup$
You will need a numerical method, e.g. Newton -Raphson method
$endgroup$
– Dr. Sonnhard Graubner
Mar 26 at 16:35
$begingroup$
We find $$xapprox 5.42225066756636430174*10^-6$$
$endgroup$
– Dr. Sonnhard Graubner
Mar 26 at 16:36
1
1
$begingroup$
I’m pretty sure that many of these types of equations can not be solved algebraically. This isn’t to say that your solution does not have one, just to say it is not guaranteed to have one.
$endgroup$
– H Huang
Mar 26 at 16:29
$begingroup$
I’m pretty sure that many of these types of equations can not be solved algebraically. This isn’t to say that your solution does not have one, just to say it is not guaranteed to have one.
$endgroup$
– H Huang
Mar 26 at 16:29
$begingroup$
Alright, thanks for the answer.
$endgroup$
– Steinar Sindri Agnarsson
Mar 26 at 16:31
$begingroup$
Alright, thanks for the answer.
$endgroup$
– Steinar Sindri Agnarsson
Mar 26 at 16:31
$begingroup$
I am writing an answer that suggests how we can analytically find a root. Should I post it?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 26 at 16:32
$begingroup$
I am writing an answer that suggests how we can analytically find a root. Should I post it?
$endgroup$
– астон вілла олоф мэллбэрг
Mar 26 at 16:32
$begingroup$
You will need a numerical method, e.g. Newton -Raphson method
$endgroup$
– Dr. Sonnhard Graubner
Mar 26 at 16:35
$begingroup$
You will need a numerical method, e.g. Newton -Raphson method
$endgroup$
– Dr. Sonnhard Graubner
Mar 26 at 16:35
$begingroup$
We find $$xapprox 5.42225066756636430174*10^-6$$
$endgroup$
– Dr. Sonnhard Graubner
Mar 26 at 16:36
$begingroup$
We find $$xapprox 5.42225066756636430174*10^-6$$
$endgroup$
– Dr. Sonnhard Graubner
Mar 26 at 16:36
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
I am pretty sure that it is impossible to solve for the equation algebraically. Let $f(x) = 2x^2 - 5x +64log x+776$. We are trying to find when $f(x) = 0$. Note that the domain of $f$ is $(0,infty)$ because of the logarithm( which I assume is the natural logarithm i.e. to the base $e$).
Going for it analytically, first we note that the derivative is $f' = 4x + frac64x - 5 = frac4x^2 - 5x+64x$ which is strictly positive on $(0,infty)$ since the numerator has discriminant negative , hence is always positive, while the denominator is always positive.
Also, from the fact that $f(x) to -infty$ as $x to 0^+$ and $f(1) >0$, we see that any root is between $0$ and $1$.
We can actually do better : note that if $x = e^y$, where $y < 0$ since we now want $0<x<1$, then we have : $$2x^2 - 5x = 2(x^2 - 2.5x) = 2((x-1.25)^2) - frac258$$
Therefore, we see that on $[0,1]$ that this part is between $0$ and $-3$.
So if equality holds, then in particular, $64y + 776$ is between $0$ and $3$, so we get $frac-77664 leq y leq frac-77364$, which after exponentiation gives $ 5.422leq 10^6x leq 5.682$. At least we get the number of significant zeros and the first significant non-zero digit, using at most two exponential operations.
$endgroup$
$begingroup$
Thanks for a great explanation!
$endgroup$
– Steinar Sindri Agnarsson
Mar 26 at 16:59
$begingroup$
You are welcome! Since you are new, up vote any answers you like, and accept the best answer so you can close the question.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 26 at 17:01
add a comment |
$begingroup$
first the derivation of this equation is greater than zero
$y=2x^2-5x+64log(x)+776$ so $fracdydx=4x-5+frac64x=frac4x^2-5x+64x$ in $4x^2-5x+64geq 0$ since $Delta leq 0$ so always $fracdydx>0$
also $y(0^+)=-infty$ and there are some $ygeq 0$ so you have a one root and no more. (since $y$ is increasing and exist $a$ and $b$ such that $f(a)f(b)<0$و Intermediate value theorem)
R code
y<-function(x)
2*x^2-5*x+64*log(x)+776
uniroot(y,c(5.414414e-06,5.423423e-06))
$root
[1] 5.423423e-06
$f.root
[1] 0.01383579
$iter
[1] 0
$init.it
[1] NA
$estim.prec
[1] 9.009e-09
so you sure have a unique answer
$endgroup$
$begingroup$
better answer use uniroot(y,lower =5.1e-06 , upper = 5.6e-06,tol=10^(-30))
$endgroup$
– masoud
Mar 26 at 17:16
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
I am pretty sure that it is impossible to solve for the equation algebraically. Let $f(x) = 2x^2 - 5x +64log x+776$. We are trying to find when $f(x) = 0$. Note that the domain of $f$ is $(0,infty)$ because of the logarithm( which I assume is the natural logarithm i.e. to the base $e$).
Going for it analytically, first we note that the derivative is $f' = 4x + frac64x - 5 = frac4x^2 - 5x+64x$ which is strictly positive on $(0,infty)$ since the numerator has discriminant negative , hence is always positive, while the denominator is always positive.
Also, from the fact that $f(x) to -infty$ as $x to 0^+$ and $f(1) >0$, we see that any root is between $0$ and $1$.
We can actually do better : note that if $x = e^y$, where $y < 0$ since we now want $0<x<1$, then we have : $$2x^2 - 5x = 2(x^2 - 2.5x) = 2((x-1.25)^2) - frac258$$
Therefore, we see that on $[0,1]$ that this part is between $0$ and $-3$.
So if equality holds, then in particular, $64y + 776$ is between $0$ and $3$, so we get $frac-77664 leq y leq frac-77364$, which after exponentiation gives $ 5.422leq 10^6x leq 5.682$. At least we get the number of significant zeros and the first significant non-zero digit, using at most two exponential operations.
$endgroup$
$begingroup$
Thanks for a great explanation!
$endgroup$
– Steinar Sindri Agnarsson
Mar 26 at 16:59
$begingroup$
You are welcome! Since you are new, up vote any answers you like, and accept the best answer so you can close the question.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 26 at 17:01
add a comment |
$begingroup$
I am pretty sure that it is impossible to solve for the equation algebraically. Let $f(x) = 2x^2 - 5x +64log x+776$. We are trying to find when $f(x) = 0$. Note that the domain of $f$ is $(0,infty)$ because of the logarithm( which I assume is the natural logarithm i.e. to the base $e$).
Going for it analytically, first we note that the derivative is $f' = 4x + frac64x - 5 = frac4x^2 - 5x+64x$ which is strictly positive on $(0,infty)$ since the numerator has discriminant negative , hence is always positive, while the denominator is always positive.
Also, from the fact that $f(x) to -infty$ as $x to 0^+$ and $f(1) >0$, we see that any root is between $0$ and $1$.
We can actually do better : note that if $x = e^y$, where $y < 0$ since we now want $0<x<1$, then we have : $$2x^2 - 5x = 2(x^2 - 2.5x) = 2((x-1.25)^2) - frac258$$
Therefore, we see that on $[0,1]$ that this part is between $0$ and $-3$.
So if equality holds, then in particular, $64y + 776$ is between $0$ and $3$, so we get $frac-77664 leq y leq frac-77364$, which after exponentiation gives $ 5.422leq 10^6x leq 5.682$. At least we get the number of significant zeros and the first significant non-zero digit, using at most two exponential operations.
$endgroup$
$begingroup$
Thanks for a great explanation!
$endgroup$
– Steinar Sindri Agnarsson
Mar 26 at 16:59
$begingroup$
You are welcome! Since you are new, up vote any answers you like, and accept the best answer so you can close the question.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 26 at 17:01
add a comment |
$begingroup$
I am pretty sure that it is impossible to solve for the equation algebraically. Let $f(x) = 2x^2 - 5x +64log x+776$. We are trying to find when $f(x) = 0$. Note that the domain of $f$ is $(0,infty)$ because of the logarithm( which I assume is the natural logarithm i.e. to the base $e$).
Going for it analytically, first we note that the derivative is $f' = 4x + frac64x - 5 = frac4x^2 - 5x+64x$ which is strictly positive on $(0,infty)$ since the numerator has discriminant negative , hence is always positive, while the denominator is always positive.
Also, from the fact that $f(x) to -infty$ as $x to 0^+$ and $f(1) >0$, we see that any root is between $0$ and $1$.
We can actually do better : note that if $x = e^y$, where $y < 0$ since we now want $0<x<1$, then we have : $$2x^2 - 5x = 2(x^2 - 2.5x) = 2((x-1.25)^2) - frac258$$
Therefore, we see that on $[0,1]$ that this part is between $0$ and $-3$.
So if equality holds, then in particular, $64y + 776$ is between $0$ and $3$, so we get $frac-77664 leq y leq frac-77364$, which after exponentiation gives $ 5.422leq 10^6x leq 5.682$. At least we get the number of significant zeros and the first significant non-zero digit, using at most two exponential operations.
$endgroup$
I am pretty sure that it is impossible to solve for the equation algebraically. Let $f(x) = 2x^2 - 5x +64log x+776$. We are trying to find when $f(x) = 0$. Note that the domain of $f$ is $(0,infty)$ because of the logarithm( which I assume is the natural logarithm i.e. to the base $e$).
Going for it analytically, first we note that the derivative is $f' = 4x + frac64x - 5 = frac4x^2 - 5x+64x$ which is strictly positive on $(0,infty)$ since the numerator has discriminant negative , hence is always positive, while the denominator is always positive.
Also, from the fact that $f(x) to -infty$ as $x to 0^+$ and $f(1) >0$, we see that any root is between $0$ and $1$.
We can actually do better : note that if $x = e^y$, where $y < 0$ since we now want $0<x<1$, then we have : $$2x^2 - 5x = 2(x^2 - 2.5x) = 2((x-1.25)^2) - frac258$$
Therefore, we see that on $[0,1]$ that this part is between $0$ and $-3$.
So if equality holds, then in particular, $64y + 776$ is between $0$ and $3$, so we get $frac-77664 leq y leq frac-77364$, which after exponentiation gives $ 5.422leq 10^6x leq 5.682$. At least we get the number of significant zeros and the first significant non-zero digit, using at most two exponential operations.
answered Mar 26 at 16:53
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
40.6k33678
$begingroup$
Thanks for a great explanation!
$endgroup$
– Steinar Sindri Agnarsson
Mar 26 at 16:59
$begingroup$
You are welcome! Since you are new, up vote any answers you like, and accept the best answer so you can close the question.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 26 at 17:01
add a comment |
$begingroup$
Thanks for a great explanation!
$endgroup$
– Steinar Sindri Agnarsson
Mar 26 at 16:59
$begingroup$
You are welcome! Since you are new, up vote any answers you like, and accept the best answer so you can close the question.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 26 at 17:01
$begingroup$
Thanks for a great explanation!
$endgroup$
– Steinar Sindri Agnarsson
Mar 26 at 16:59
$begingroup$
Thanks for a great explanation!
$endgroup$
– Steinar Sindri Agnarsson
Mar 26 at 16:59
$begingroup$
You are welcome! Since you are new, up vote any answers you like, and accept the best answer so you can close the question.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 26 at 17:01
$begingroup$
You are welcome! Since you are new, up vote any answers you like, and accept the best answer so you can close the question.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 26 at 17:01
add a comment |
$begingroup$
first the derivation of this equation is greater than zero
$y=2x^2-5x+64log(x)+776$ so $fracdydx=4x-5+frac64x=frac4x^2-5x+64x$ in $4x^2-5x+64geq 0$ since $Delta leq 0$ so always $fracdydx>0$
also $y(0^+)=-infty$ and there are some $ygeq 0$ so you have a one root and no more. (since $y$ is increasing and exist $a$ and $b$ such that $f(a)f(b)<0$و Intermediate value theorem)
R code
y<-function(x)
2*x^2-5*x+64*log(x)+776
uniroot(y,c(5.414414e-06,5.423423e-06))
$root
[1] 5.423423e-06
$f.root
[1] 0.01383579
$iter
[1] 0
$init.it
[1] NA
$estim.prec
[1] 9.009e-09
so you sure have a unique answer
$endgroup$
$begingroup$
better answer use uniroot(y,lower =5.1e-06 , upper = 5.6e-06,tol=10^(-30))
$endgroup$
– masoud
Mar 26 at 17:16
add a comment |
$begingroup$
first the derivation of this equation is greater than zero
$y=2x^2-5x+64log(x)+776$ so $fracdydx=4x-5+frac64x=frac4x^2-5x+64x$ in $4x^2-5x+64geq 0$ since $Delta leq 0$ so always $fracdydx>0$
also $y(0^+)=-infty$ and there are some $ygeq 0$ so you have a one root and no more. (since $y$ is increasing and exist $a$ and $b$ such that $f(a)f(b)<0$و Intermediate value theorem)
R code
y<-function(x)
2*x^2-5*x+64*log(x)+776
uniroot(y,c(5.414414e-06,5.423423e-06))
$root
[1] 5.423423e-06
$f.root
[1] 0.01383579
$iter
[1] 0
$init.it
[1] NA
$estim.prec
[1] 9.009e-09
so you sure have a unique answer
$endgroup$
$begingroup$
better answer use uniroot(y,lower =5.1e-06 , upper = 5.6e-06,tol=10^(-30))
$endgroup$
– masoud
Mar 26 at 17:16
add a comment |
$begingroup$
first the derivation of this equation is greater than zero
$y=2x^2-5x+64log(x)+776$ so $fracdydx=4x-5+frac64x=frac4x^2-5x+64x$ in $4x^2-5x+64geq 0$ since $Delta leq 0$ so always $fracdydx>0$
also $y(0^+)=-infty$ and there are some $ygeq 0$ so you have a one root and no more. (since $y$ is increasing and exist $a$ and $b$ such that $f(a)f(b)<0$و Intermediate value theorem)
R code
y<-function(x)
2*x^2-5*x+64*log(x)+776
uniroot(y,c(5.414414e-06,5.423423e-06))
$root
[1] 5.423423e-06
$f.root
[1] 0.01383579
$iter
[1] 0
$init.it
[1] NA
$estim.prec
[1] 9.009e-09
so you sure have a unique answer
$endgroup$
first the derivation of this equation is greater than zero
$y=2x^2-5x+64log(x)+776$ so $fracdydx=4x-5+frac64x=frac4x^2-5x+64x$ in $4x^2-5x+64geq 0$ since $Delta leq 0$ so always $fracdydx>0$
also $y(0^+)=-infty$ and there are some $ygeq 0$ so you have a one root and no more. (since $y$ is increasing and exist $a$ and $b$ such that $f(a)f(b)<0$و Intermediate value theorem)
R code
y<-function(x)
2*x^2-5*x+64*log(x)+776
uniroot(y,c(5.414414e-06,5.423423e-06))
$root
[1] 5.423423e-06
$f.root
[1] 0.01383579
$iter
[1] 0
$init.it
[1] NA
$estim.prec
[1] 9.009e-09
so you sure have a unique answer
edited Mar 26 at 16:51
answered Mar 26 at 16:46
masoudmasoud
32218
32218
$begingroup$
better answer use uniroot(y,lower =5.1e-06 , upper = 5.6e-06,tol=10^(-30))
$endgroup$
– masoud
Mar 26 at 17:16
add a comment |
$begingroup$
better answer use uniroot(y,lower =5.1e-06 , upper = 5.6e-06,tol=10^(-30))
$endgroup$
– masoud
Mar 26 at 17:16
$begingroup$
better answer use uniroot(y,lower =5.1e-06 , upper = 5.6e-06,tol=10^(-30))
$endgroup$
– masoud
Mar 26 at 17:16
$begingroup$
better answer use uniroot(y,lower =5.1e-06 , upper = 5.6e-06,tol=10^(-30))
$endgroup$
– masoud
Mar 26 at 17:16
add a comment |
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$begingroup$
I’m pretty sure that many of these types of equations can not be solved algebraically. This isn’t to say that your solution does not have one, just to say it is not guaranteed to have one.
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– H Huang
Mar 26 at 16:29
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Alright, thanks for the answer.
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– Steinar Sindri Agnarsson
Mar 26 at 16:31
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I am writing an answer that suggests how we can analytically find a root. Should I post it?
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– астон вілла олоф мэллбэрг
Mar 26 at 16:32
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You will need a numerical method, e.g. Newton -Raphson method
$endgroup$
– Dr. Sonnhard Graubner
Mar 26 at 16:35
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We find $$xapprox 5.42225066756636430174*10^-6$$
$endgroup$
– Dr. Sonnhard Graubner
Mar 26 at 16:36