Prove that two functions grow at equal rate
$begingroup$
Prove that functions $g(x)=ln(ln(x))$ and $h(x)=ln(lg(x))$ grow at equal rate for every base and value of x.
I'm actually very confused about what 'for every base' actually means. I'm assuming that I'm supposed to keep the outside function as the same, but the inside can be a logarithm with any base?
I can solve this by differentiating both functions, but then again I'm not sure if that's the way to go since it probably doesn't explain the base part. Sorry, if I didn't explain this very well.
logarithms
edited Nov 29 '18 at 15:33
user376343
3,3282825
asked Nov 29 '18 at 15:32
mathisharddmathishardd
1
$endgroup$
add a comment |
$begingroup$
Prove that functions $g(x)=ln(ln(x))$ and $h(x)=ln(lg(x))$ grow at equal rate for every base and value of x.
I'm actually very confused about what 'for every base' actually means. I'm assuming that I'm supposed to keep the outside function as the same, but the inside can be a logarithm with any base?
I can solve this by differentiating both functions, but then again I'm not sure if that's the way to go since it probably doesn't explain the base part. Sorry, if I didn't explain this very well.
logarithms
edited Nov 29 '18 at 15:33
user376343
3,3282825
asked Nov 29 '18 at 15:32
mathisharddmathishardd
1
$endgroup$
$begingroup$
I think it as you are writing: $lg = log_a$ for any $a$ convenient as a basis of a logarithme .... but there are basis such that $log_a$ is decreasing (!)
$endgroup$
– user376343
Nov 29 '18 at 15:36
$begingroup$
You need to get the question right. You probably mean that they grow at equal rate (look up your definition) as a function of $x$ for every $a$ that is the base of the $lg$ function.
$endgroup$
– Ross Millikan
Nov 29 '18 at 15:45
add a comment |
$begingroup$
Prove that functions $g(x)=ln(ln(x))$ and $h(x)=ln(lg(x))$ grow at equal rate for every base and value of x.
I'm actually very confused about what 'for every base' actually means. I'm assuming that I'm supposed to keep the outside function as the same, but the inside can be a logarithm with any base?
I can solve this by differentiating both functions, but then again I'm not sure if that's the way to go since it probably doesn't explain the base part. Sorry, if I didn't explain this very well.
logarithms
edited Nov 29 '18 at 15:33
user376343
3,3282825
asked Nov 29 '18 at 15:32
mathisharddmathishardd
1
$endgroup$
Prove that functions $g(x)=ln(ln(x))$ and $h(x)=ln(lg(x))$ grow at equal rate for every base and value of x.
I'm actually very confused about what 'for every base' actually means. I'm assuming that I'm supposed to keep the outside function as the same, but the inside can be a logarithm with any base?
I can solve this by differentiating both functions, but then again I'm not sure if that's the way to go since it probably doesn't explain the base part. Sorry, if I didn't explain this very well.
logarithms
logarithms
edited Nov 29 '18 at 15:33
user376343
3,3282825
asked Nov 29 '18 at 15:32
mathisharddmathishardd
1
edited Nov 29 '18 at 15:33
user376343
3,3282825
asked Nov 29 '18 at 15:32
mathisharddmathishardd
1
edited Nov 29 '18 at 15:33
user376343
3,3282825
edited Nov 29 '18 at 15:33
user376343
3,3282825
edited Nov 29 '18 at 15:33
user376343
3,3282825
3,3282825
asked Nov 29 '18 at 15:32
mathisharddmathishardd
1
asked Nov 29 '18 at 15:32
mathisharddmathishardd
1
asked Nov 29 '18 at 15:32
mathisharddmathishardd
1
1
$begingroup$
I think it as you are writing: $lg = log_a$ for any $a$ convenient as a basis of a logarithme .... but there are basis such that $log_a$ is decreasing (!)
$endgroup$
– user376343
Nov 29 '18 at 15:36
$begingroup$
You need to get the question right. You probably mean that they grow at equal rate (look up your definition) as a function of $x$ for every $a$ that is the base of the $lg$ function.
$endgroup$
– Ross Millikan
Nov 29 '18 at 15:45
add a comment |
$begingroup$
I think it as you are writing: $lg = log_a$ for any $a$ convenient as a basis of a logarithme .... but there are basis such that $log_a$ is decreasing (!)
$endgroup$
– user376343
Nov 29 '18 at 15:36
$begingroup$
You need to get the question right. You probably mean that they grow at equal rate (look up your definition) as a function of $x$ for every $a$ that is the base of the $lg$ function.
$endgroup$
– Ross Millikan
Nov 29 '18 at 15:45
$begingroup$
I think it as you are writing: $lg = log_a$ for any $a$ convenient as a basis of a logarithme .... but there are basis such that $log_a$ is decreasing (!)
$endgroup$
– user376343
Nov 29 '18 at 15:36
$begingroup$
I think it as you are writing: $lg = log_a$ for any $a$ convenient as a basis of a logarithme .... but there are basis such that $log_a$ is decreasing (!)
$endgroup$
– user376343
Nov 29 '18 at 15:36
$begingroup$
You need to get the question right. You probably mean that they grow at equal rate (look up your definition) as a function of $x$ for every $a$ that is the base of the $lg$ function.
$endgroup$
– Ross Millikan
Nov 29 '18 at 15:45
$begingroup$
You need to get the question right. You probably mean that they grow at equal rate (look up your definition) as a function of $x$ for every $a$ that is the base of the $lg$ function.
$endgroup$
– Ross Millikan
Nov 29 '18 at 15:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Presumably your $lg(x)$ is $log_a(x)$. I believe you are to assume $a,x gt 1$.
Now use the laws of logarithms to express $log_a(x)$ in terms of $ln(x)$, then pull out the correction term.
answered Nov 29 '18 at 15:47
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
Another hint, assuming that your lg$(x)$ is $log_{a}(x)$. Note that $log_{a}(x)=ln(x)/ln(a)$...
answered Nov 29 '18 at 15:54
FrobeniusFrobenius
613
$endgroup$
$begingroup$
Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
$endgroup$
– mathishardd
Nov 29 '18 at 16:00
1
$begingroup$
Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
$endgroup$
– Frobenius
Nov 29 '18 at 16:05
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Hint: Presumably your $lg(x)$ is $log_a(x)$. I believe you are to assume $a,x gt 1$.
Now use the laws of logarithms to express $log_a(x)$ in terms of $ln(x)$, then pull out the correction term.
answered Nov 29 '18 at 15:47
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
Hint: Presumably your $lg(x)$ is $log_a(x)$. I believe you are to assume $a,x gt 1$.
Now use the laws of logarithms to express $log_a(x)$ in terms of $ln(x)$, then pull out the correction term.
answered Nov 29 '18 at 15:47
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
Hint: Presumably your $lg(x)$ is $log_a(x)$. I believe you are to assume $a,x gt 1$.
Now use the laws of logarithms to express $log_a(x)$ in terms of $ln(x)$, then pull out the correction term.
answered Nov 29 '18 at 15:47
Ross MillikanRoss Millikan
293k23197371
$endgroup$
Hint: Presumably your $lg(x)$ is $log_a(x)$. I believe you are to assume $a,x gt 1$.
Now use the laws of logarithms to express $log_a(x)$ in terms of $ln(x)$, then pull out the correction term.
answered Nov 29 '18 at 15:47
Ross MillikanRoss Millikan
293k23197371
answered Nov 29 '18 at 15:47
Ross MillikanRoss Millikan
293k23197371
answered Nov 29 '18 at 15:47
Ross MillikanRoss Millikan
293k23197371
answered Nov 29 '18 at 15:47
Ross MillikanRoss Millikan
293k23197371
293k23197371
add a comment |
add a comment |
$begingroup$
Another hint, assuming that your lg$(x)$ is $log_{a}(x)$. Note that $log_{a}(x)=ln(x)/ln(a)$...
answered Nov 29 '18 at 15:54
FrobeniusFrobenius
613
$endgroup$
$begingroup$
Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
$endgroup$
– mathishardd
Nov 29 '18 at 16:00
1
$begingroup$
Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
$endgroup$
– Frobenius
Nov 29 '18 at 16:05
add a comment |
$begingroup$
Another hint, assuming that your lg$(x)$ is $log_{a}(x)$. Note that $log_{a}(x)=ln(x)/ln(a)$...
answered Nov 29 '18 at 15:54
FrobeniusFrobenius
613
$endgroup$
$begingroup$
Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
$endgroup$
– mathishardd
Nov 29 '18 at 16:00
1
$begingroup$
Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
$endgroup$
– Frobenius
Nov 29 '18 at 16:05
add a comment |
$begingroup$
Another hint, assuming that your lg$(x)$ is $log_{a}(x)$. Note that $log_{a}(x)=ln(x)/ln(a)$...
answered Nov 29 '18 at 15:54
FrobeniusFrobenius
613
$endgroup$
Another hint, assuming that your lg$(x)$ is $log_{a}(x)$. Note that $log_{a}(x)=ln(x)/ln(a)$...
answered Nov 29 '18 at 15:54
FrobeniusFrobenius
613
answered Nov 29 '18 at 15:54
FrobeniusFrobenius
613
answered Nov 29 '18 at 15:54
FrobeniusFrobenius
613
answered Nov 29 '18 at 15:54
FrobeniusFrobenius
613
613
$begingroup$
Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
$endgroup$
– mathishardd
Nov 29 '18 at 16:00
1
$begingroup$
Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
$endgroup$
– Frobenius
Nov 29 '18 at 16:05
add a comment |
$begingroup$
Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
$endgroup$
– mathishardd
Nov 29 '18 at 16:00
1
$begingroup$
Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
$endgroup$
– Frobenius
Nov 29 '18 at 16:05
$begingroup$
Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
$endgroup$
– mathishardd
Nov 29 '18 at 16:00
$begingroup$
Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
$endgroup$
– mathishardd
Nov 29 '18 at 16:00
1
1
$begingroup$
Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
$endgroup$
– Frobenius
Nov 29 '18 at 16:05
$begingroup$
Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
$endgroup$
– Frobenius
Nov 29 '18 at 16:05
add a comment |
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$begingroup$
I think it as you are writing: $lg = log_a$ for any $a$ convenient as a basis of a logarithme .... but there are basis such that $log_a$ is decreasing (!)
$endgroup$
– user376343
Nov 29 '18 at 15:36
$begingroup$
You need to get the question right. You probably mean that they grow at equal rate (look up your definition) as a function of $x$ for every $a$ that is the base of the $lg$ function.
$endgroup$
– Ross Millikan
Nov 29 '18 at 15:45