Prove that the product of the roots of $x^{log_{2016}x}*sqrt{2016}=x^{2016}$ is a Natural Number [duplicate]
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This question already has an answer here:
The product of the two roots of $sqrt{2014}x^{log_{2014} x}=x^{2014}$ is an integer. Find its units digit
4 answers
Prove that the product of the roots of $x^{log_{2016}x}*sqrt{2016}=x^{2016}$ is a Natural Number.
This is my solution:
by putting $log_{2016}$ on both sides we get:
$log^2{_{2016}}x-2016log_{2016}x=-log_{2016}sqrt{2016}$
then by putting $log_{2016}x$ in front of bracket on left side, then removing $log_{2016}$ from both sides and squaring the equation, and $t=log_{2016}x$
$t^2-4032t+2016^2-2016=0$
solving for $t_{1/2}$ we get:
$x_1 = 2016^{12left(168+sqrt{14}right)}$
$x_2 = 2016^{12left(168-sqrt{14}right)}$
$x_1 * x_2 = 2016^{4032}$
Is my solution correct? (I think I might have a mistake because I haven't solved this type of problem before, and overall I just started practicing). And if correct is there a better/easier way to solve it?
algebra-precalculus
asked Nov 30 '18 at 22:24
PeroPero
1497
$endgroup$
marked as duplicate by астон вілла олоф мэллбэрг, Community♦ Nov 30 '18 at 23:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
The product of the two roots of $sqrt{2014}x^{log_{2014} x}=x^{2014}$ is an integer. Find its units digit
4 answers
Prove that the product of the roots of $x^{log_{2016}x}*sqrt{2016}=x^{2016}$ is a Natural Number.
This is my solution:
by putting $log_{2016}$ on both sides we get:
$log^2{_{2016}}x-2016log_{2016}x=-log_{2016}sqrt{2016}$
then by putting $log_{2016}x$ in front of bracket on left side, then removing $log_{2016}$ from both sides and squaring the equation, and $t=log_{2016}x$
$t^2-4032t+2016^2-2016=0$
solving for $t_{1/2}$ we get:
$x_1 = 2016^{12left(168+sqrt{14}right)}$
$x_2 = 2016^{12left(168-sqrt{14}right)}$
$x_1 * x_2 = 2016^{4032}$
Is my solution correct? (I think I might have a mistake because I haven't solved this type of problem before, and overall I just started practicing). And if correct is there a better/easier way to solve it?
algebra-precalculus
asked Nov 30 '18 at 22:24
PeroPero
1497
$endgroup$
marked as duplicate by астон вілла олоф мэллбэрг, Community♦ Nov 30 '18 at 23:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
this question is a near duplicate.
$endgroup$
– lulu
Nov 30 '18 at 22:25
$begingroup$
Oh it actually is the same. My bad I guess. I searched for my problem and nothing came up. That link is the exact same problem but with 2014 instead of 2016.
$endgroup$
– Pero
Nov 30 '18 at 22:29
$begingroup$
Do I delete my post now? Or learn to solve it from the link and answer my question? Or what?
$endgroup$
– Pero
Nov 30 '18 at 22:30
$begingroup$
Sounds fine to me!
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 22:31
$begingroup$
Leave your question as is. We will close it as a duplicate. I have added one above.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 23:09
add a comment |
$begingroup$
This question already has an answer here:
The product of the two roots of $sqrt{2014}x^{log_{2014} x}=x^{2014}$ is an integer. Find its units digit
4 answers
Prove that the product of the roots of $x^{log_{2016}x}*sqrt{2016}=x^{2016}$ is a Natural Number.
This is my solution:
by putting $log_{2016}$ on both sides we get:
$log^2{_{2016}}x-2016log_{2016}x=-log_{2016}sqrt{2016}$
then by putting $log_{2016}x$ in front of bracket on left side, then removing $log_{2016}$ from both sides and squaring the equation, and $t=log_{2016}x$
$t^2-4032t+2016^2-2016=0$
solving for $t_{1/2}$ we get:
$x_1 = 2016^{12left(168+sqrt{14}right)}$
$x_2 = 2016^{12left(168-sqrt{14}right)}$
$x_1 * x_2 = 2016^{4032}$
Is my solution correct? (I think I might have a mistake because I haven't solved this type of problem before, and overall I just started practicing). And if correct is there a better/easier way to solve it?
algebra-precalculus
asked Nov 30 '18 at 22:24
PeroPero
1497
$endgroup$
This question already has an answer here:
The product of the two roots of $sqrt{2014}x^{log_{2014} x}=x^{2014}$ is an integer. Find its units digit
4 answers
Prove that the product of the roots of $x^{log_{2016}x}*sqrt{2016}=x^{2016}$ is a Natural Number.
This is my solution:
by putting $log_{2016}$ on both sides we get:
$log^2{_{2016}}x-2016log_{2016}x=-log_{2016}sqrt{2016}$
then by putting $log_{2016}x$ in front of bracket on left side, then removing $log_{2016}$ from both sides and squaring the equation, and $t=log_{2016}x$
$t^2-4032t+2016^2-2016=0$
solving for $t_{1/2}$ we get:
$x_1 = 2016^{12left(168+sqrt{14}right)}$
$x_2 = 2016^{12left(168-sqrt{14}right)}$
$x_1 * x_2 = 2016^{4032}$
Is my solution correct? (I think I might have a mistake because I haven't solved this type of problem before, and overall I just started practicing). And if correct is there a better/easier way to solve it?
This question already has an answer here:
The product of the two roots of $sqrt{2014}x^{log_{2014} x}=x^{2014}$ is an integer. Find its units digit
4 answers
algebra-precalculus
algebra-precalculus
asked Nov 30 '18 at 22:24
PeroPero
1497
asked Nov 30 '18 at 22:24
PeroPero
1497
asked Nov 30 '18 at 22:24
PeroPero
1497
asked Nov 30 '18 at 22:24
PeroPero
1497
asked Nov 30 '18 at 22:24
PeroPero
1497
1497
marked as duplicate by астон вілла олоф мэллбэрг, Community♦ Nov 30 '18 at 23:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by астон вілла олоф мэллбэрг, Community♦ Nov 30 '18 at 23:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
this question is a near duplicate.
$endgroup$
– lulu
Nov 30 '18 at 22:25
$begingroup$
Oh it actually is the same. My bad I guess. I searched for my problem and nothing came up. That link is the exact same problem but with 2014 instead of 2016.
$endgroup$
– Pero
Nov 30 '18 at 22:29
$begingroup$
Do I delete my post now? Or learn to solve it from the link and answer my question? Or what?
$endgroup$
– Pero
Nov 30 '18 at 22:30
$begingroup$
Sounds fine to me!
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 22:31
$begingroup$
Leave your question as is. We will close it as a duplicate. I have added one above.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 23:09
add a comment |
$begingroup$
this question is a near duplicate.
$endgroup$
– lulu
Nov 30 '18 at 22:25
$begingroup$
Oh it actually is the same. My bad I guess. I searched for my problem and nothing came up. That link is the exact same problem but with 2014 instead of 2016.
$endgroup$
– Pero
Nov 30 '18 at 22:29
$begingroup$
Do I delete my post now? Or learn to solve it from the link and answer my question? Or what?
$endgroup$
– Pero
Nov 30 '18 at 22:30
$begingroup$
Sounds fine to me!
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 22:31
$begingroup$
Leave your question as is. We will close it as a duplicate. I have added one above.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 23:09
$begingroup$
this question is a near duplicate.
$endgroup$
– lulu
Nov 30 '18 at 22:25
$begingroup$
this question is a near duplicate.
$endgroup$
– lulu
Nov 30 '18 at 22:25
$begingroup$
Oh it actually is the same. My bad I guess. I searched for my problem and nothing came up. That link is the exact same problem but with 2014 instead of 2016.
$endgroup$
– Pero
Nov 30 '18 at 22:29
$begingroup$
Oh it actually is the same. My bad I guess. I searched for my problem and nothing came up. That link is the exact same problem but with 2014 instead of 2016.
$endgroup$
– Pero
Nov 30 '18 at 22:29
$begingroup$
Do I delete my post now? Or learn to solve it from the link and answer my question? Or what?
$endgroup$
– Pero
Nov 30 '18 at 22:30
$begingroup$
Do I delete my post now? Or learn to solve it from the link and answer my question? Or what?
$endgroup$
– Pero
Nov 30 '18 at 22:30
$begingroup$
Sounds fine to me!
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 22:31
$begingroup$
Sounds fine to me!
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 22:31
$begingroup$
Leave your question as is. We will close it as a duplicate. I have added one above.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 23:09
$begingroup$
Leave your question as is. We will close it as a duplicate. I have added one above.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 23:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We have that
$$y=log_{2016}x iff 2016^y=x implies x^y=2016^{y^2}$$
$$x^{log_{2016}x}cdot sqrt{2016}=x^{2016} iff 2016^{y^2}cdot sqrt{2016}=2016^{2016y}$$
$$2016^{(y^2-2016y)} =frac{1}{sqrt{2016}} iff y^2-2016y=frac12$$
answered Nov 30 '18 at 22:31
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
Apparently, I could've solved it a lot easier. Thanks to @lulu for providing the link.
$x^{log_{2016}x}*sqrt{2016}=x^{2016}$
By taking $log_{2016}$ of both sides we get:
$log^2{_{2016}}x-2016log_{2016}x+1/2=0$
$t^2-2016t+1/2=0$
Let r, s be the roots of the original problem, we have:
$log_{2016} rs = log_{2016} r + log_{2016} s = 2016 => rs = 2016^{2016}$
answered Nov 30 '18 at 23:11
PeroPero
1497
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have that
$$y=log_{2016}x iff 2016^y=x implies x^y=2016^{y^2}$$
$$x^{log_{2016}x}cdot sqrt{2016}=x^{2016} iff 2016^{y^2}cdot sqrt{2016}=2016^{2016y}$$
$$2016^{(y^2-2016y)} =frac{1}{sqrt{2016}} iff y^2-2016y=frac12$$
answered Nov 30 '18 at 22:31
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
We have that
$$y=log_{2016}x iff 2016^y=x implies x^y=2016^{y^2}$$
$$x^{log_{2016}x}cdot sqrt{2016}=x^{2016} iff 2016^{y^2}cdot sqrt{2016}=2016^{2016y}$$
$$2016^{(y^2-2016y)} =frac{1}{sqrt{2016}} iff y^2-2016y=frac12$$
answered Nov 30 '18 at 22:31
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
We have that
$$y=log_{2016}x iff 2016^y=x implies x^y=2016^{y^2}$$
$$x^{log_{2016}x}cdot sqrt{2016}=x^{2016} iff 2016^{y^2}cdot sqrt{2016}=2016^{2016y}$$
$$2016^{(y^2-2016y)} =frac{1}{sqrt{2016}} iff y^2-2016y=frac12$$
answered Nov 30 '18 at 22:31
gimusigimusi
92.8k84494
$endgroup$
We have that
$$y=log_{2016}x iff 2016^y=x implies x^y=2016^{y^2}$$
$$x^{log_{2016}x}cdot sqrt{2016}=x^{2016} iff 2016^{y^2}cdot sqrt{2016}=2016^{2016y}$$
$$2016^{(y^2-2016y)} =frac{1}{sqrt{2016}} iff y^2-2016y=frac12$$
answered Nov 30 '18 at 22:31
gimusigimusi
92.8k84494
answered Nov 30 '18 at 22:31
gimusigimusi
92.8k84494
answered Nov 30 '18 at 22:31
gimusigimusi
92.8k84494
answered Nov 30 '18 at 22:31
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
$begingroup$
Apparently, I could've solved it a lot easier. Thanks to @lulu for providing the link.
$x^{log_{2016}x}*sqrt{2016}=x^{2016}$
By taking $log_{2016}$ of both sides we get:
$log^2{_{2016}}x-2016log_{2016}x+1/2=0$
$t^2-2016t+1/2=0$
Let r, s be the roots of the original problem, we have:
$log_{2016} rs = log_{2016} r + log_{2016} s = 2016 => rs = 2016^{2016}$
answered Nov 30 '18 at 23:11
PeroPero
1497
$endgroup$
add a comment |
$begingroup$
Apparently, I could've solved it a lot easier. Thanks to @lulu for providing the link.
$x^{log_{2016}x}*sqrt{2016}=x^{2016}$
By taking $log_{2016}$ of both sides we get:
$log^2{_{2016}}x-2016log_{2016}x+1/2=0$
$t^2-2016t+1/2=0$
Let r, s be the roots of the original problem, we have:
$log_{2016} rs = log_{2016} r + log_{2016} s = 2016 => rs = 2016^{2016}$
answered Nov 30 '18 at 23:11
PeroPero
1497
$endgroup$
add a comment |
$begingroup$
Apparently, I could've solved it a lot easier. Thanks to @lulu for providing the link.
$x^{log_{2016}x}*sqrt{2016}=x^{2016}$
By taking $log_{2016}$ of both sides we get:
$log^2{_{2016}}x-2016log_{2016}x+1/2=0$
$t^2-2016t+1/2=0$
Let r, s be the roots of the original problem, we have:
$log_{2016} rs = log_{2016} r + log_{2016} s = 2016 => rs = 2016^{2016}$
answered Nov 30 '18 at 23:11
PeroPero
1497
$endgroup$
Apparently, I could've solved it a lot easier. Thanks to @lulu for providing the link.
$x^{log_{2016}x}*sqrt{2016}=x^{2016}$
By taking $log_{2016}$ of both sides we get:
$log^2{_{2016}}x-2016log_{2016}x+1/2=0$
$t^2-2016t+1/2=0$
Let r, s be the roots of the original problem, we have:
$log_{2016} rs = log_{2016} r + log_{2016} s = 2016 => rs = 2016^{2016}$
answered Nov 30 '18 at 23:11
PeroPero
1497
answered Nov 30 '18 at 23:11
PeroPero
1497
answered Nov 30 '18 at 23:11
PeroPero
1497
answered Nov 30 '18 at 23:11
PeroPero
1497
1497
add a comment |
add a comment |
$begingroup$
this question is a near duplicate.
$endgroup$
– lulu
Nov 30 '18 at 22:25
$begingroup$
Oh it actually is the same. My bad I guess. I searched for my problem and nothing came up. That link is the exact same problem but with 2014 instead of 2016.
$endgroup$
– Pero
Nov 30 '18 at 22:29
$begingroup$
Do I delete my post now? Or learn to solve it from the link and answer my question? Or what?
$endgroup$
– Pero
Nov 30 '18 at 22:30
$begingroup$
Sounds fine to me!
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 22:31
$begingroup$
Leave your question as is. We will close it as a duplicate. I have added one above.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 23:09