Mixing
which one is larger? 8 factorial to the power of 1/8 or 9 factorial to the power of 1/9
$begingroup$
Which one is larger?
8 factorial to the power of 1/8 or 9 factorial to the power of 1/9. Show your work without calculator.
arithmetic
asked Jan 21 at 11:20
KevinKevin
143
$endgroup$
add a comment |
$begingroup$
Which one is larger?
8 factorial to the power of 1/8 or 9 factorial to the power of 1/9. Show your work without calculator.
arithmetic
asked Jan 21 at 11:20
KevinKevin
143
$endgroup$
4
$begingroup$
As the exercise requires: "Show your work." :-)
$endgroup$
– trancelocation
Jan 21 at 11:23
$begingroup$
sry 4 replying late,actly,at first, ttl have no clue working on this question. @trancelocation
$endgroup$
– Kevin
Jan 21 at 13:19
add a comment |
$begingroup$
Which one is larger?
8 factorial to the power of 1/8 or 9 factorial to the power of 1/9. Show your work without calculator.
arithmetic
asked Jan 21 at 11:20
KevinKevin
143
$endgroup$
Which one is larger?
8 factorial to the power of 1/8 or 9 factorial to the power of 1/9. Show your work without calculator.
arithmetic
arithmetic
asked Jan 21 at 11:20
KevinKevin
143
asked Jan 21 at 11:20
KevinKevin
143
asked Jan 21 at 11:20
KevinKevin
143
asked Jan 21 at 11:20
KevinKevin
143
asked Jan 21 at 11:20
KevinKevin
143
143
4
$begingroup$
As the exercise requires: "Show your work." :-)
$endgroup$
– trancelocation
Jan 21 at 11:23
$begingroup$
sry 4 replying late,actly,at first, ttl have no clue working on this question. @trancelocation
$endgroup$
– Kevin
Jan 21 at 13:19
add a comment |
4
$begingroup$
As the exercise requires: "Show your work." :-)
$endgroup$
– trancelocation
Jan 21 at 11:23
$begingroup$
sry 4 replying late,actly,at first, ttl have no clue working on this question. @trancelocation
$endgroup$
– Kevin
Jan 21 at 13:19
4
4
$begingroup$
As the exercise requires: "Show your work." :-)
$endgroup$
– trancelocation
Jan 21 at 11:23
$begingroup$
As the exercise requires: "Show your work." :-)
$endgroup$
– trancelocation
Jan 21 at 11:23
$begingroup$
sry 4 replying late,actly,at first, ttl have no clue working on this question. @trancelocation
$endgroup$
– Kevin
Jan 21 at 13:19
$begingroup$
sry 4 replying late,actly,at first, ttl have no clue working on this question. @trancelocation
$endgroup$
– Kevin
Jan 21 at 13:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Compare $(8!)^9=(8!)(8!)^8$ and $(9!)^8=9^8(8!)^8$.
answered Jan 21 at 11:23
lhflhf
166k10171396
$endgroup$
$begingroup$
thx! since 8! is less than 9^8, (8!)^9 is less than (9!)^8, thus both them to the power of 1/72 then we have 8 factorial to the power of 1/8 is less than 9 factorial to the power of 1/9.
$endgroup$
– Kevin
Jan 21 at 11:32
add a comment |
$begingroup$
Define the function $f(x) = x^{frac{1}{x}}$ and calculate its derivative. You should get $f'(x)=-x^{frac{1}{x}-2}(ln(x)-1)$. You see that that $forall x > e: f'(x) < 0$ since $ln(e)=1$. Thus, the function $f(x)$ is strictly decreasing function for $x>e approx 2.718$. Hence, $f(8) > f(9)$.
answered Jan 21 at 11:48
SelosSelos
539
$endgroup$
1
$begingroup$
You need $f(x)=(x!)^{1/x}$. And probably use $Gamma$ instead of factorial.
$endgroup$
– lhf
Jan 21 at 12:03
$begingroup$
True, I missed the factorial part. It's still possible but your solution seems more elegant, faster and easier.
$endgroup$
– Selos
Jan 21 at 12:15
add a comment |
Your Answer
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2 Answers
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active
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2 Answers
2
active
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active
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votes
$begingroup$
Hint: Compare $(8!)^9=(8!)(8!)^8$ and $(9!)^8=9^8(8!)^8$.
answered Jan 21 at 11:23
lhflhf
166k10171396
$endgroup$
$begingroup$
thx! since 8! is less than 9^8, (8!)^9 is less than (9!)^8, thus both them to the power of 1/72 then we have 8 factorial to the power of 1/8 is less than 9 factorial to the power of 1/9.
$endgroup$
– Kevin
Jan 21 at 11:32
add a comment |
$begingroup$
Hint: Compare $(8!)^9=(8!)(8!)^8$ and $(9!)^8=9^8(8!)^8$.
answered Jan 21 at 11:23
lhflhf
166k10171396
$endgroup$
$begingroup$
thx! since 8! is less than 9^8, (8!)^9 is less than (9!)^8, thus both them to the power of 1/72 then we have 8 factorial to the power of 1/8 is less than 9 factorial to the power of 1/9.
$endgroup$
– Kevin
Jan 21 at 11:32
add a comment |
$begingroup$
Hint: Compare $(8!)^9=(8!)(8!)^8$ and $(9!)^8=9^8(8!)^8$.
answered Jan 21 at 11:23
lhflhf
166k10171396
$endgroup$
Hint: Compare $(8!)^9=(8!)(8!)^8$ and $(9!)^8=9^8(8!)^8$.
answered Jan 21 at 11:23
lhflhf
166k10171396
edited Jan 21 at 11:42
answered Jan 21 at 11:23
lhflhf
166k10171396
answered Jan 21 at 11:23
lhflhf
166k10171396
answered Jan 21 at 11:23
lhflhf
166k10171396
166k10171396
$begingroup$
thx! since 8! is less than 9^8, (8!)^9 is less than (9!)^8, thus both them to the power of 1/72 then we have 8 factorial to the power of 1/8 is less than 9 factorial to the power of 1/9.
$endgroup$
– Kevin
Jan 21 at 11:32
add a comment |
$begingroup$
thx! since 8! is less than 9^8, (8!)^9 is less than (9!)^8, thus both them to the power of 1/72 then we have 8 factorial to the power of 1/8 is less than 9 factorial to the power of 1/9.
$endgroup$
– Kevin
Jan 21 at 11:32
$begingroup$
thx! since 8! is less than 9^8, (8!)^9 is less than (9!)^8, thus both them to the power of 1/72 then we have 8 factorial to the power of 1/8 is less than 9 factorial to the power of 1/9.
$endgroup$
– Kevin
Jan 21 at 11:32
$begingroup$
thx! since 8! is less than 9^8, (8!)^9 is less than (9!)^8, thus both them to the power of 1/72 then we have 8 factorial to the power of 1/8 is less than 9 factorial to the power of 1/9.
$endgroup$
– Kevin
Jan 21 at 11:32
add a comment |
$begingroup$
Define the function $f(x) = x^{frac{1}{x}}$ and calculate its derivative. You should get $f'(x)=-x^{frac{1}{x}-2}(ln(x)-1)$. You see that that $forall x > e: f'(x) < 0$ since $ln(e)=1$. Thus, the function $f(x)$ is strictly decreasing function for $x>e approx 2.718$. Hence, $f(8) > f(9)$.
answered Jan 21 at 11:48
SelosSelos
539
$endgroup$
1
$begingroup$
You need $f(x)=(x!)^{1/x}$. And probably use $Gamma$ instead of factorial.
$endgroup$
– lhf
Jan 21 at 12:03
$begingroup$
True, I missed the factorial part. It's still possible but your solution seems more elegant, faster and easier.
$endgroup$
– Selos
Jan 21 at 12:15
add a comment |
$begingroup$
Define the function $f(x) = x^{frac{1}{x}}$ and calculate its derivative. You should get $f'(x)=-x^{frac{1}{x}-2}(ln(x)-1)$. You see that that $forall x > e: f'(x) < 0$ since $ln(e)=1$. Thus, the function $f(x)$ is strictly decreasing function for $x>e approx 2.718$. Hence, $f(8) > f(9)$.
answered Jan 21 at 11:48
SelosSelos
539
$endgroup$
1
$begingroup$
You need $f(x)=(x!)^{1/x}$. And probably use $Gamma$ instead of factorial.
$endgroup$
– lhf
Jan 21 at 12:03
$begingroup$
True, I missed the factorial part. It's still possible but your solution seems more elegant, faster and easier.
$endgroup$
– Selos
Jan 21 at 12:15
add a comment |
$begingroup$
Define the function $f(x) = x^{frac{1}{x}}$ and calculate its derivative. You should get $f'(x)=-x^{frac{1}{x}-2}(ln(x)-1)$. You see that that $forall x > e: f'(x) < 0$ since $ln(e)=1$. Thus, the function $f(x)$ is strictly decreasing function for $x>e approx 2.718$. Hence, $f(8) > f(9)$.
answered Jan 21 at 11:48
SelosSelos
539
$endgroup$
Define the function $f(x) = x^{frac{1}{x}}$ and calculate its derivative. You should get $f'(x)=-x^{frac{1}{x}-2}(ln(x)-1)$. You see that that $forall x > e: f'(x) < 0$ since $ln(e)=1$. Thus, the function $f(x)$ is strictly decreasing function for $x>e approx 2.718$. Hence, $f(8) > f(9)$.
answered Jan 21 at 11:48
SelosSelos
539
answered Jan 21 at 11:48
SelosSelos
539
answered Jan 21 at 11:48
SelosSelos
539
answered Jan 21 at 11:48
SelosSelos
539
539
1
$begingroup$
You need $f(x)=(x!)^{1/x}$. And probably use $Gamma$ instead of factorial.
$endgroup$
– lhf
Jan 21 at 12:03
$begingroup$
True, I missed the factorial part. It's still possible but your solution seems more elegant, faster and easier.
$endgroup$
– Selos
Jan 21 at 12:15
add a comment |
1
$begingroup$
You need $f(x)=(x!)^{1/x}$. And probably use $Gamma$ instead of factorial.
$endgroup$
– lhf
Jan 21 at 12:03
$begingroup$
True, I missed the factorial part. It's still possible but your solution seems more elegant, faster and easier.
$endgroup$
– Selos
Jan 21 at 12:15
1
1
$begingroup$
You need $f(x)=(x!)^{1/x}$. And probably use $Gamma$ instead of factorial.
$endgroup$
– lhf
Jan 21 at 12:03
$begingroup$
You need $f(x)=(x!)^{1/x}$. And probably use $Gamma$ instead of factorial.
$endgroup$
– lhf
Jan 21 at 12:03
$begingroup$
True, I missed the factorial part. It's still possible but your solution seems more elegant, faster and easier.
$endgroup$
– Selos
Jan 21 at 12:15
$begingroup$
True, I missed the factorial part. It's still possible but your solution seems more elegant, faster and easier.
$endgroup$
– Selos
Jan 21 at 12:15
add a comment |
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4
$begingroup$
As the exercise requires: "Show your work." :-)
$endgroup$
– trancelocation
Jan 21 at 11:23
$begingroup$
sry 4 replying late,actly,at first, ttl have no clue working on this question. @trancelocation
$endgroup$
– Kevin
Jan 21 at 13:19