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Prove: $log_2(x)+log_x(y)+log_y(8)geq sqrt[3]81$
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up vote
2
down vote
favorite
Prove that for every $x$,$y$ greater than $1$:
$$log_2(x)+log_x(y)+log_y(8)geq sqrt[3]81$$
What I've tried has got me to:
$$fraclog_y(x)log_y(2)+log_x(y)+3log_y(2)geq sqrt[3]81$$
I didn't really get far.. I can't see where I can go from here, especially not what to do with $ sqrt[3]81$.
This is taken out of the maths entry tests for TAU, so this shouldn't be too hard.
algebra-precalculus inequality logarithms a.m.-g.m.-inequality
edited Jul 16 at 19:11
TheSimpliFire
9,70861951
asked Jul 16 at 19:01
Maxim
616314
add a comment |Â
up vote
2
down vote
favorite
Prove that for every $x$,$y$ greater than $1$:
$$log_2(x)+log_x(y)+log_y(8)geq sqrt[3]81$$
What I've tried has got me to:
$$fraclog_y(x)log_y(2)+log_x(y)+3log_y(2)geq sqrt[3]81$$
I didn't really get far.. I can't see where I can go from here, especially not what to do with $ sqrt[3]81$.
This is taken out of the maths entry tests for TAU, so this shouldn't be too hard.
algebra-precalculus inequality logarithms a.m.-g.m.-inequality
edited Jul 16 at 19:11
TheSimpliFire
9,70861951
asked Jul 16 at 19:01
Maxim
616314
$log_xy=frac1log_yx$
– MalayTheDynamo
Jul 16 at 19:04
A neat thing about $log_a b = frac 1log_b a$ and $frac log_b clog_b a = log_a c$.... It means $log_a b* log_b c = log_a c$ and so $log_a b*log_b c* log_c d*....... *log_y z = log_a z$......
– fleablood
Jul 16 at 19:40
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Prove that for every $x$,$y$ greater than $1$:
$$log_2(x)+log_x(y)+log_y(8)geq sqrt[3]81$$
What I've tried has got me to:
$$fraclog_y(x)log_y(2)+log_x(y)+3log_y(2)geq sqrt[3]81$$
I didn't really get far.. I can't see where I can go from here, especially not what to do with $ sqrt[3]81$.
This is taken out of the maths entry tests for TAU, so this shouldn't be too hard.
algebra-precalculus inequality logarithms a.m.-g.m.-inequality
edited Jul 16 at 19:11
TheSimpliFire
9,70861951
asked Jul 16 at 19:01
Maxim
616314
Prove that for every $x$,$y$ greater than $1$:
$$log_2(x)+log_x(y)+log_y(8)geq sqrt[3]81$$
What I've tried has got me to:
$$fraclog_y(x)log_y(2)+log_x(y)+3log_y(2)geq sqrt[3]81$$
I didn't really get far.. I can't see where I can go from here, especially not what to do with $ sqrt[3]81$.
This is taken out of the maths entry tests for TAU, so this shouldn't be too hard.
algebra-precalculus inequality logarithms a.m.-g.m.-inequality
edited Jul 16 at 19:11
TheSimpliFire
9,70861951
asked Jul 16 at 19:01
Maxim
616314
edited Jul 16 at 19:11
TheSimpliFire
9,70861951
edited Jul 16 at 19:11
TheSimpliFire
9,70861951
edited Jul 16 at 19:11
TheSimpliFire
9,70861951
9,70861951
asked Jul 16 at 19:01
Maxim
616314
asked Jul 16 at 19:01
Maxim
616314
asked Jul 16 at 19:01
Maxim
616314
616314
$log_xy=frac1log_yx$
– MalayTheDynamo
Jul 16 at 19:04
A neat thing about $log_a b = frac 1log_b a$ and $frac log_b clog_b a = log_a c$.... It means $log_a b* log_b c = log_a c$ and so $log_a b*log_b c* log_c d*....... *log_y z = log_a z$......
– fleablood
Jul 16 at 19:40
add a comment |Â
$log_xy=frac1log_yx$
– MalayTheDynamo
Jul 16 at 19:04
A neat thing about $log_a b = frac 1log_b a$ and $frac log_b clog_b a = log_a c$.... It means $log_a b* log_b c = log_a c$ and so $log_a b*log_b c* log_c d*....... *log_y z = log_a z$......
– fleablood
Jul 16 at 19:40
$log_xy=frac1log_yx$
– MalayTheDynamo
Jul 16 at 19:04
$log_xy=frac1log_yx$
– MalayTheDynamo
Jul 16 at 19:04
A neat thing about $log_a b = frac 1log_b a$ and $frac log_b clog_b a = log_a c$.... It means $log_a b* log_b c = log_a c$ and so $log_a b*log_b c* log_c d*....... *log_y z = log_a z$......
– fleablood
Jul 16 at 19:40
A neat thing about $log_a b = frac 1log_b a$ and $frac log_b clog_b a = log_a c$.... It means $log_a b* log_b c = log_a c$ and so $log_a b*log_b c* log_c d*....... *log_y z = log_a z$......
– fleablood
Jul 16 at 19:40
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
We have $$log_2x+log_xy+log_y8=fraclog xlog2+fraclog ylog x+frac3log2log y$$
and since $2,x,y>1$, we can deduce that their logarithms are non-negative, and so will their quotients.
Now use AM-GM:
$$sqrt[3]fraclog xlog2cdotfraclog ylog xcdotfrac3log2log ylefracfraclog xlog2+fraclog ylog x+frac3log2log y3$$ giving $$fraclog xlog2+fraclog ylog x+frac3log2log yge3sqrt[3]3=sqrt[3]81$$ as required.
answered Jul 16 at 19:10
TheSimpliFire
9,70861951
While I did all these logarithmic exercises, I noticed that the AM-GM is actually a very common technique for the solution. Is there any reason for that or is it just a coincudence?
– Maxim
Jul 17 at 12:41
@Maxim AM-GM is usually the second inequality that people learn (after $x^2 ge 0$). So it's a good choice to use when writing a problem for an entrance exam, because many people are familiar with it.
– Ovi
Jul 17 at 17:37
@Ovi That's embarrassing. I learned about AM/GM/HM just last week :)
– TheSimpliFire
Jul 17 at 19:22
@TheSimpliFire Don't feel embarrased, I'm not saying that most people know about and how to use AM-GM. I'm saying that of the people who know inequalities, most know AM-GM; so if they put an inequality question, AM-GM is a good problem. However, the question of weather they should put inequalities questions is separate.
– Ovi
Jul 17 at 21:39
add a comment |Â
up vote
5
down vote
Let's convert everything to $log$ base $2$ so we have a common something to work with:
$$log_2 x + dfrac log_2 ylog_2 x + dfrac log_2 8log_2 y ge sqrt [3]81 = 3 sqrt[3]3$$
Now let $a = log_2 x$ and $b = log_2 y$. The condition $x, y > 1$ implies that $a, b > 0$. So now we have to prove
$$a + dfrac ab + dfrac 3b ge3 sqrt[3]3$$
The cube root in there especially may remind us of AM-GM with $3$ terms. And indeed, the inequality $textAMleft(a, dfrac ab, dfrac 3bright) ge textGMleft(a, dfrac ab, dfrac 3bright)$ gives exactly what is desired.
answered Jul 16 at 19:11
Ovi
11.3k935105
add a comment |Â
up vote
2
down vote
Hint: Show that $a+fracba+frac3bgeq sqrt[3]81$ for all $a,b>0$, using AM-GM.
answered Jul 16 at 19:05
Batominovski
23.2k22777
add a comment |Â
up vote
0
down vote
Two things:
$log_a b = frac 1log_b a$ and $frac log_b clog_b a = log_a c$ so $log_a blog_b c = frac log_b clog_b a = log_a c$.
And AM-GM says $frac a + b+ c3 ge sqrt[3]abc$.
So.....
$frac log_2 x + log_x y + log_y 83 ge sqrt[3]log_2 x log_x y log_y 8 = sqrt[3]log_2 8$
answered Jul 16 at 19:37
fleablood
60.5k22575
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
We have $$log_2x+log_xy+log_y8=fraclog xlog2+fraclog ylog x+frac3log2log y$$
and since $2,x,y>1$, we can deduce that their logarithms are non-negative, and so will their quotients.
Now use AM-GM:
$$sqrt[3]fraclog xlog2cdotfraclog ylog xcdotfrac3log2log ylefracfraclog xlog2+fraclog ylog x+frac3log2log y3$$ giving $$fraclog xlog2+fraclog ylog x+frac3log2log yge3sqrt[3]3=sqrt[3]81$$ as required.
answered Jul 16 at 19:10
TheSimpliFire
9,70861951
While I did all these logarithmic exercises, I noticed that the AM-GM is actually a very common technique for the solution. Is there any reason for that or is it just a coincudence?
– Maxim
Jul 17 at 12:41
@Maxim AM-GM is usually the second inequality that people learn (after $x^2 ge 0$). So it's a good choice to use when writing a problem for an entrance exam, because many people are familiar with it.
– Ovi
Jul 17 at 17:37
@Ovi That's embarrassing. I learned about AM/GM/HM just last week :)
– TheSimpliFire
Jul 17 at 19:22
@TheSimpliFire Don't feel embarrased, I'm not saying that most people know about and how to use AM-GM. I'm saying that of the people who know inequalities, most know AM-GM; so if they put an inequality question, AM-GM is a good problem. However, the question of weather they should put inequalities questions is separate.
– Ovi
Jul 17 at 21:39
add a comment |Â
up vote
3
down vote
accepted
We have $$log_2x+log_xy+log_y8=fraclog xlog2+fraclog ylog x+frac3log2log y$$
and since $2,x,y>1$, we can deduce that their logarithms are non-negative, and so will their quotients.
Now use AM-GM:
$$sqrt[3]fraclog xlog2cdotfraclog ylog xcdotfrac3log2log ylefracfraclog xlog2+fraclog ylog x+frac3log2log y3$$ giving $$fraclog xlog2+fraclog ylog x+frac3log2log yge3sqrt[3]3=sqrt[3]81$$ as required.
answered Jul 16 at 19:10
TheSimpliFire
9,70861951
While I did all these logarithmic exercises, I noticed that the AM-GM is actually a very common technique for the solution. Is there any reason for that or is it just a coincudence?
– Maxim
Jul 17 at 12:41
@Maxim AM-GM is usually the second inequality that people learn (after $x^2 ge 0$). So it's a good choice to use when writing a problem for an entrance exam, because many people are familiar with it.
– Ovi
Jul 17 at 17:37
@Ovi That's embarrassing. I learned about AM/GM/HM just last week :)
– TheSimpliFire
Jul 17 at 19:22
@TheSimpliFire Don't feel embarrased, I'm not saying that most people know about and how to use AM-GM. I'm saying that of the people who know inequalities, most know AM-GM; so if they put an inequality question, AM-GM is a good problem. However, the question of weather they should put inequalities questions is separate.
– Ovi
Jul 17 at 21:39
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
We have $$log_2x+log_xy+log_y8=fraclog xlog2+fraclog ylog x+frac3log2log y$$
and since $2,x,y>1$, we can deduce that their logarithms are non-negative, and so will their quotients.
Now use AM-GM:
$$sqrt[3]fraclog xlog2cdotfraclog ylog xcdotfrac3log2log ylefracfraclog xlog2+fraclog ylog x+frac3log2log y3$$ giving $$fraclog xlog2+fraclog ylog x+frac3log2log yge3sqrt[3]3=sqrt[3]81$$ as required.
answered Jul 16 at 19:10
TheSimpliFire
9,70861951
We have $$log_2x+log_xy+log_y8=fraclog xlog2+fraclog ylog x+frac3log2log y$$
and since $2,x,y>1$, we can deduce that their logarithms are non-negative, and so will their quotients.
Now use AM-GM:
$$sqrt[3]fraclog xlog2cdotfraclog ylog xcdotfrac3log2log ylefracfraclog xlog2+fraclog ylog x+frac3log2log y3$$ giving $$fraclog xlog2+fraclog ylog x+frac3log2log yge3sqrt[3]3=sqrt[3]81$$ as required.
answered Jul 16 at 19:10
TheSimpliFire
9,70861951
edited Jul 16 at 19:15
answered Jul 16 at 19:10
TheSimpliFire
9,70861951
answered Jul 16 at 19:10
TheSimpliFire
9,70861951
answered Jul 16 at 19:10
TheSimpliFire
9,70861951
9,70861951
While I did all these logarithmic exercises, I noticed that the AM-GM is actually a very common technique for the solution. Is there any reason for that or is it just a coincudence?
– Maxim
Jul 17 at 12:41
@Maxim AM-GM is usually the second inequality that people learn (after $x^2 ge 0$). So it's a good choice to use when writing a problem for an entrance exam, because many people are familiar with it.
– Ovi
Jul 17 at 17:37
@Ovi That's embarrassing. I learned about AM/GM/HM just last week :)
– TheSimpliFire
Jul 17 at 19:22
@TheSimpliFire Don't feel embarrased, I'm not saying that most people know about and how to use AM-GM. I'm saying that of the people who know inequalities, most know AM-GM; so if they put an inequality question, AM-GM is a good problem. However, the question of weather they should put inequalities questions is separate.
– Ovi
Jul 17 at 21:39
add a comment |Â
While I did all these logarithmic exercises, I noticed that the AM-GM is actually a very common technique for the solution. Is there any reason for that or is it just a coincudence?
– Maxim
Jul 17 at 12:41
@Maxim AM-GM is usually the second inequality that people learn (after $x^2 ge 0$). So it's a good choice to use when writing a problem for an entrance exam, because many people are familiar with it.
– Ovi
Jul 17 at 17:37
@Ovi That's embarrassing. I learned about AM/GM/HM just last week :)
– TheSimpliFire
Jul 17 at 19:22
@TheSimpliFire Don't feel embarrased, I'm not saying that most people know about and how to use AM-GM. I'm saying that of the people who know inequalities, most know AM-GM; so if they put an inequality question, AM-GM is a good problem. However, the question of weather they should put inequalities questions is separate.
– Ovi
Jul 17 at 21:39
While I did all these logarithmic exercises, I noticed that the AM-GM is actually a very common technique for the solution. Is there any reason for that or is it just a coincudence?
– Maxim
Jul 17 at 12:41
While I did all these logarithmic exercises, I noticed that the AM-GM is actually a very common technique for the solution. Is there any reason for that or is it just a coincudence?
– Maxim
Jul 17 at 12:41
@Maxim AM-GM is usually the second inequality that people learn (after $x^2 ge 0$). So it's a good choice to use when writing a problem for an entrance exam, because many people are familiar with it.
– Ovi
Jul 17 at 17:37
@Maxim AM-GM is usually the second inequality that people learn (after $x^2 ge 0$). So it's a good choice to use when writing a problem for an entrance exam, because many people are familiar with it.
– Ovi
Jul 17 at 17:37
@Ovi That's embarrassing. I learned about AM/GM/HM just last week :)
– TheSimpliFire
Jul 17 at 19:22
@Ovi That's embarrassing. I learned about AM/GM/HM just last week :)
– TheSimpliFire
Jul 17 at 19:22
@TheSimpliFire Don't feel embarrased, I'm not saying that most people know about and how to use AM-GM. I'm saying that of the people who know inequalities, most know AM-GM; so if they put an inequality question, AM-GM is a good problem. However, the question of weather they should put inequalities questions is separate.
– Ovi
Jul 17 at 21:39
@TheSimpliFire Don't feel embarrased, I'm not saying that most people know about and how to use AM-GM. I'm saying that of the people who know inequalities, most know AM-GM; so if they put an inequality question, AM-GM is a good problem. However, the question of weather they should put inequalities questions is separate.
– Ovi
Jul 17 at 21:39
add a comment |Â
up vote
5
down vote
Let's convert everything to $log$ base $2$ so we have a common something to work with:
$$log_2 x + dfrac log_2 ylog_2 x + dfrac log_2 8log_2 y ge sqrt [3]81 = 3 sqrt[3]3$$
Now let $a = log_2 x$ and $b = log_2 y$. The condition $x, y > 1$ implies that $a, b > 0$. So now we have to prove
$$a + dfrac ab + dfrac 3b ge3 sqrt[3]3$$
The cube root in there especially may remind us of AM-GM with $3$ terms. And indeed, the inequality $textAMleft(a, dfrac ab, dfrac 3bright) ge textGMleft(a, dfrac ab, dfrac 3bright)$ gives exactly what is desired.
answered Jul 16 at 19:11
Ovi
11.3k935105
add a comment |Â
up vote
5
down vote
Let's convert everything to $log$ base $2$ so we have a common something to work with:
$$log_2 x + dfrac log_2 ylog_2 x + dfrac log_2 8log_2 y ge sqrt [3]81 = 3 sqrt[3]3$$
Now let $a = log_2 x$ and $b = log_2 y$. The condition $x, y > 1$ implies that $a, b > 0$. So now we have to prove
$$a + dfrac ab + dfrac 3b ge3 sqrt[3]3$$
The cube root in there especially may remind us of AM-GM with $3$ terms. And indeed, the inequality $textAMleft(a, dfrac ab, dfrac 3bright) ge textGMleft(a, dfrac ab, dfrac 3bright)$ gives exactly what is desired.
answered Jul 16 at 19:11
Ovi
11.3k935105
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Let's convert everything to $log$ base $2$ so we have a common something to work with:
$$log_2 x + dfrac log_2 ylog_2 x + dfrac log_2 8log_2 y ge sqrt [3]81 = 3 sqrt[3]3$$
Now let $a = log_2 x$ and $b = log_2 y$. The condition $x, y > 1$ implies that $a, b > 0$. So now we have to prove
$$a + dfrac ab + dfrac 3b ge3 sqrt[3]3$$
The cube root in there especially may remind us of AM-GM with $3$ terms. And indeed, the inequality $textAMleft(a, dfrac ab, dfrac 3bright) ge textGMleft(a, dfrac ab, dfrac 3bright)$ gives exactly what is desired.
answered Jul 16 at 19:11
Ovi
11.3k935105
Let's convert everything to $log$ base $2$ so we have a common something to work with:
$$log_2 x + dfrac log_2 ylog_2 x + dfrac log_2 8log_2 y ge sqrt [3]81 = 3 sqrt[3]3$$
Now let $a = log_2 x$ and $b = log_2 y$. The condition $x, y > 1$ implies that $a, b > 0$. So now we have to prove
$$a + dfrac ab + dfrac 3b ge3 sqrt[3]3$$
The cube root in there especially may remind us of AM-GM with $3$ terms. And indeed, the inequality $textAMleft(a, dfrac ab, dfrac 3bright) ge textGMleft(a, dfrac ab, dfrac 3bright)$ gives exactly what is desired.
answered Jul 16 at 19:11
Ovi
11.3k935105
edited Jul 16 at 19:15
answered Jul 16 at 19:11
Ovi
11.3k935105
answered Jul 16 at 19:11
Ovi
11.3k935105
answered Jul 16 at 19:11
Ovi
11.3k935105
11.3k935105
add a comment |Â
add a comment |Â
up vote
2
down vote
Hint: Show that $a+fracba+frac3bgeq sqrt[3]81$ for all $a,b>0$, using AM-GM.
answered Jul 16 at 19:05
Batominovski
23.2k22777
add a comment |Â
up vote
2
down vote
Hint: Show that $a+fracba+frac3bgeq sqrt[3]81$ for all $a,b>0$, using AM-GM.
answered Jul 16 at 19:05
Batominovski
23.2k22777
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: Show that $a+fracba+frac3bgeq sqrt[3]81$ for all $a,b>0$, using AM-GM.
answered Jul 16 at 19:05
Batominovski
23.2k22777
Hint: Show that $a+fracba+frac3bgeq sqrt[3]81$ for all $a,b>0$, using AM-GM.
answered Jul 16 at 19:05
Batominovski
23.2k22777
answered Jul 16 at 19:05
Batominovski
23.2k22777
answered Jul 16 at 19:05
Batominovski
23.2k22777
answered Jul 16 at 19:05
Batominovski
23.2k22777
23.2k22777
add a comment |Â
add a comment |Â
up vote
0
down vote
Two things:
$log_a b = frac 1log_b a$ and $frac log_b clog_b a = log_a c$ so $log_a blog_b c = frac log_b clog_b a = log_a c$.
And AM-GM says $frac a + b+ c3 ge sqrt[3]abc$.
So.....
$frac log_2 x + log_x y + log_y 83 ge sqrt[3]log_2 x log_x y log_y 8 = sqrt[3]log_2 8$
answered Jul 16 at 19:37
fleablood
60.5k22575
add a comment |Â
up vote
0
down vote
Two things:
$log_a b = frac 1log_b a$ and $frac log_b clog_b a = log_a c$ so $log_a blog_b c = frac log_b clog_b a = log_a c$.
And AM-GM says $frac a + b+ c3 ge sqrt[3]abc$.
So.....
$frac log_2 x + log_x y + log_y 83 ge sqrt[3]log_2 x log_x y log_y 8 = sqrt[3]log_2 8$
answered Jul 16 at 19:37
fleablood
60.5k22575
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Two things:
$log_a b = frac 1log_b a$ and $frac log_b clog_b a = log_a c$ so $log_a blog_b c = frac log_b clog_b a = log_a c$.
And AM-GM says $frac a + b+ c3 ge sqrt[3]abc$.
So.....
$frac log_2 x + log_x y + log_y 83 ge sqrt[3]log_2 x log_x y log_y 8 = sqrt[3]log_2 8$
answered Jul 16 at 19:37
fleablood
60.5k22575
Two things:
$log_a b = frac 1log_b a$ and $frac log_b clog_b a = log_a c$ so $log_a blog_b c = frac log_b clog_b a = log_a c$.
And AM-GM says $frac a + b+ c3 ge sqrt[3]abc$.
So.....
$frac log_2 x + log_x y + log_y 83 ge sqrt[3]log_2 x log_x y log_y 8 = sqrt[3]log_2 8$
answered Jul 16 at 19:37
fleablood
60.5k22575
edited Jul 16 at 19:45
answered Jul 16 at 19:37
fleablood
60.5k22575
answered Jul 16 at 19:37
fleablood
60.5k22575
answered Jul 16 at 19:37
fleablood
60.5k22575
60.5k22575
add a comment |Â
add a comment |Â
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$log_xy=frac1log_yx$
– MalayTheDynamo
Jul 16 at 19:04
A neat thing about $log_a b = frac 1log_b a$ and $frac log_b clog_b a = log_a c$.... It means $log_a b* log_b c = log_a c$ and so $log_a b*log_b c* log_c d*....... *log_y z = log_a z$......
– fleablood
Jul 16 at 19:40