How do I find the solutions of $|x-2|^10x^2-1=|x-2|^3x$? The Next CEO of Stack OverflowSum of all real numbers $x$ such that $(textA quadratic)^textAnother quadratic=1$.How to find solutions of this equation?How to find the roots of $x^4 +1$Find all real roots of $x^5+10x^3+20x - 4=0$Find real solutions for a mod equation with powerComplex solutions to $ x^3 + 512 = 0 $finding integer solutions for a and bFind the real solutions of the equations:$x+y=2$, $xy-z^2=1$How to manually calculate/approximate a specific value of the Lambert W functionSolving $x^4 -10x^3 + 26x^2 -10x +1 = 0$.Why do some logarithmic equations have two solutions?
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How do I find the solutions of $|x-2|^10x^2-1=|x-2|^3x$?
The Next CEO of Stack OverflowSum of all real numbers $x$ such that $(textA quadratic)^textAnother quadratic=1$.How to find solutions of this equation?How to find the roots of $x^4 +1$Find all real roots of $x^5+10x^3+20x - 4=0$Find real solutions for a mod equation with powerComplex solutions to $ x^3 + 512 = 0 $finding integer solutions for a and bFind the real solutions of the equations:$x+y=2$, $xy-z^2=1$How to manually calculate/approximate a specific value of the Lambert W functionSolving $x^4 -10x^3 + 26x^2 -10x +1 = 0$.Why do some logarithmic equations have two solutions?
$begingroup$
How do I find the solutions of the following equation: $$|x-2|^10x^2-1=|x-2|^3x?$$
It has 5 solutions, 4 positive and 1 negative. The graphs are these.
How do I compute the values of these roots manually?
algebra-precalculus logarithms
edited 4 hours ago
Maria Mazur
48.8k1260122
asked yesterday
Namami ShankerNamami Shanker
141
New contributor
Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
How do I find the solutions of the following equation: $$|x-2|^10x^2-1=|x-2|^3x?$$
It has 5 solutions, 4 positive and 1 negative. The graphs are these.
How do I compute the values of these roots manually?
algebra-precalculus logarithms
edited 4 hours ago
Maria Mazur
48.8k1260122
asked yesterday
Namami ShankerNamami Shanker
141
New contributor
Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
yesterday
add a comment |
$begingroup$
How do I find the solutions of the following equation: $$|x-2|^10x^2-1=|x-2|^3x?$$
It has 5 solutions, 4 positive and 1 negative. The graphs are these.
How do I compute the values of these roots manually?
algebra-precalculus logarithms
edited 4 hours ago
Maria Mazur
48.8k1260122
asked yesterday
Namami ShankerNamami Shanker
141
New contributor
Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
How do I find the solutions of the following equation: $$|x-2|^10x^2-1=|x-2|^3x?$$
It has 5 solutions, 4 positive and 1 negative. The graphs are these.
How do I compute the values of these roots manually?
algebra-precalculus logarithms
algebra-precalculus logarithms
edited 4 hours ago
Maria Mazur
48.8k1260122
asked yesterday
Namami ShankerNamami Shanker
141
New contributor
Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
Maria Mazur
48.8k1260122
asked yesterday
Namami ShankerNamami Shanker
141
New contributor
Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
Maria Mazur
48.8k1260122
edited 4 hours ago
Maria Mazur
48.8k1260122
edited 4 hours ago
Maria Mazur
48.8k1260122
48.8k1260122
asked yesterday
Namami ShankerNamami Shanker
141
New contributor
Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Namami ShankerNamami Shanker
141
asked yesterday
Namami ShankerNamami Shanker
141
141
New contributor
Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
yesterday
add a comment |
$begingroup$
math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
yesterday
$begingroup$
math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
yesterday
$begingroup$
math.stackexchange.com/questions/3157637/…
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– lab bhattacharjee
yesterday
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
We see that $x=2$ is one solution. Let $xne 2$.
Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$
So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.
Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.
edited yesterday
Moo
5,64131020
answered yesterday
Maria MazurMaria Mazur
48.8k1260122
$endgroup$
add a comment |
$begingroup$
So rearranging gives
$$|x-2|^10x^2-1-|x-2|^3x=0$$
$$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.
answered yesterday
Peter ForemanPeter Foreman
4,4121216
$endgroup$
add a comment |
$begingroup$
Hint
Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
We get easy that $x=2$ is one solution. Now let $xneq 2$, then it must be $10x^2-1=3x$. Can you finish? Hint: $x=3$ and $x=1$ are also solutions.
edited 21 hours ago
Solomon Ucko
14219
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
$begingroup$
Yes thank you sir.
$endgroup$
– Namami Shanker
yesterday
$begingroup$
This does not give all of the solutions.
$endgroup$
– Peter Foreman
yesterday
$begingroup$
The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
$endgroup$
– Robert Israel
yesterday
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We see that $x=2$ is one solution. Let $xne 2$.
Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$
So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.
Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.
edited yesterday
Moo
5,64131020
answered yesterday
Maria MazurMaria Mazur
48.8k1260122
$endgroup$
add a comment |
$begingroup$
We see that $x=2$ is one solution. Let $xne 2$.
Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$
So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.
Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.
edited yesterday
Moo
5,64131020
answered yesterday
Maria MazurMaria Mazur
48.8k1260122
$endgroup$
add a comment |
$begingroup$
We see that $x=2$ is one solution. Let $xne 2$.
Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$
So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.
Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.
edited yesterday
Moo
5,64131020
answered yesterday
Maria MazurMaria Mazur
48.8k1260122
$endgroup$
We see that $x=2$ is one solution. Let $xne 2$.
Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$
So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.
Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.
edited yesterday
Moo
5,64131020
answered yesterday
Maria MazurMaria Mazur
48.8k1260122
edited yesterday
Moo
5,64131020
edited yesterday
Moo
5,64131020
edited yesterday
Moo
5,64131020
5,64131020
answered yesterday
Maria MazurMaria Mazur
48.8k1260122
answered yesterday
Maria MazurMaria Mazur
48.8k1260122
answered yesterday
Maria MazurMaria Mazur
48.8k1260122
48.8k1260122
add a comment |
add a comment |
$begingroup$
So rearranging gives
$$|x-2|^10x^2-1-|x-2|^3x=0$$
$$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.
answered yesterday
Peter ForemanPeter Foreman
4,4121216
$endgroup$
add a comment |
$begingroup$
So rearranging gives
$$|x-2|^10x^2-1-|x-2|^3x=0$$
$$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.
answered yesterday
Peter ForemanPeter Foreman
4,4121216
$endgroup$
add a comment |
$begingroup$
So rearranging gives
$$|x-2|^10x^2-1-|x-2|^3x=0$$
$$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.
answered yesterday
Peter ForemanPeter Foreman
4,4121216
$endgroup$
So rearranging gives
$$|x-2|^10x^2-1-|x-2|^3x=0$$
$$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.
answered yesterday
Peter ForemanPeter Foreman
4,4121216
answered yesterday
Peter ForemanPeter Foreman
4,4121216
answered yesterday
Peter ForemanPeter Foreman
4,4121216
answered yesterday
Peter ForemanPeter Foreman
4,4121216
4,4121216
add a comment |
add a comment |
$begingroup$
Hint
Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Hint
Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Hint
Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
Hint
Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
answered yesterday
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
add a comment |
add a comment |
$begingroup$
We get easy that $x=2$ is one solution. Now let $xneq 2$, then it must be $10x^2-1=3x$. Can you finish? Hint: $x=3$ and $x=1$ are also solutions.
edited 21 hours ago
Solomon Ucko
14219
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
$begingroup$
Yes thank you sir.
$endgroup$
– Namami Shanker
yesterday
$begingroup$
This does not give all of the solutions.
$endgroup$
– Peter Foreman
yesterday
$begingroup$
The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
$endgroup$
– Robert Israel
yesterday
add a comment |
$begingroup$
We get easy that $x=2$ is one solution. Now let $xneq 2$, then it must be $10x^2-1=3x$. Can you finish? Hint: $x=3$ and $x=1$ are also solutions.
edited 21 hours ago
Solomon Ucko
14219
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
$begingroup$
Yes thank you sir.
$endgroup$
– Namami Shanker
yesterday
$begingroup$
This does not give all of the solutions.
$endgroup$
– Peter Foreman
yesterday
$begingroup$
The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
$endgroup$
– Robert Israel
yesterday
add a comment |
$begingroup$
We get easy that $x=2$ is one solution. Now let $xneq 2$, then it must be $10x^2-1=3x$. Can you finish? Hint: $x=3$ and $x=1$ are also solutions.
edited 21 hours ago
Solomon Ucko
14219
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
We get easy that $x=2$ is one solution. Now let $xneq 2$, then it must be $10x^2-1=3x$. Can you finish? Hint: $x=3$ and $x=1$ are also solutions.
edited 21 hours ago
Solomon Ucko
14219
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
edited 21 hours ago
Solomon Ucko
14219
edited 21 hours ago
Solomon Ucko
14219
edited 21 hours ago
Solomon Ucko
14219
14219
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
78.2k42867
$begingroup$
Yes thank you sir.
$endgroup$
– Namami Shanker
yesterday
$begingroup$
This does not give all of the solutions.
$endgroup$
– Peter Foreman
yesterday
$begingroup$
The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
$endgroup$
– Robert Israel
yesterday
add a comment |
$begingroup$
Yes thank you sir.
$endgroup$
– Namami Shanker
yesterday
$begingroup$
This does not give all of the solutions.
$endgroup$
– Peter Foreman
yesterday
$begingroup$
The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
$endgroup$
– Robert Israel
yesterday
$begingroup$
Yes thank you sir.
$endgroup$
– Namami Shanker
yesterday
$begingroup$
Yes thank you sir.
$endgroup$
– Namami Shanker
yesterday
$begingroup$
This does not give all of the solutions.
$endgroup$
– Peter Foreman
yesterday
$begingroup$
This does not give all of the solutions.
$endgroup$
– Peter Foreman
yesterday
$begingroup$
The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
$endgroup$
– Robert Israel
yesterday
$begingroup$
The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
$endgroup$
– Robert Israel
yesterday
add a comment |
Namami Shanker is a new contributor. Be nice, and check out our Code of Conduct.
Namami Shanker is a new contributor. Be nice, and check out our Code of Conduct.
Namami Shanker is a new contributor. Be nice, and check out our Code of Conduct.
Namami Shanker is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
yesterday