Prove that two functions grow at equal rate
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 at 15:33
user376343
2,7782822
asked Nov 29 at 15:32
mathishardd
1
add a comment |
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 at 15:33
user376343
2,7782822
asked Nov 29 at 15:32
mathishardd
1
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 (!)
– user376343
Nov 29 at 15:36
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.
– Ross Millikan
Nov 29 at 15:45
add a comment |
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 at 15:33
user376343
2,7782822
asked Nov 29 at 15:32
mathishardd
1
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 at 15:33
user376343
2,7782822
asked Nov 29 at 15:32
mathishardd
1
edited Nov 29 at 15:33
user376343
2,7782822
asked Nov 29 at 15:32
mathishardd
1
edited Nov 29 at 15:33
user376343
2,7782822
edited Nov 29 at 15:33
user376343
2,7782822
edited Nov 29 at 15:33
user376343
2,7782822
2,7782822
asked Nov 29 at 15:32
mathishardd
1
asked Nov 29 at 15:32
mathishardd
1
asked Nov 29 at 15:32
mathishardd
1
1
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 (!)
– user376343
Nov 29 at 15:36
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.
– Ross Millikan
Nov 29 at 15:45
add a comment |
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 (!)
– user376343
Nov 29 at 15:36
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.
– Ross Millikan
Nov 29 at 15:45
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 (!)
– user376343
Nov 29 at 15:36
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 (!)
– user376343
Nov 29 at 15:36
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.
– Ross Millikan
Nov 29 at 15:45
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.
– Ross Millikan
Nov 29 at 15:45
add a comment |
2 Answers
2
active
oldest
votes
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 at 15:47
Ross Millikan
291k23196370
add a comment |
Another hint, assuming that your lg$(x)$ is $log_{a}(x)$. Note that $log_{a}(x)=ln(x)/ln(a)$...
answered Nov 29 at 15:54
Frobenius
613
Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
– mathishardd
Nov 29 at 16:00
1
Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
– Frobenius
Nov 29 at 16:05
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
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oldest
votes
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 at 15:47
Ross Millikan
291k23196370
add a comment |
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 at 15:47
Ross Millikan
291k23196370
add a comment |
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 at 15:47
Ross Millikan
291k23196370
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 at 15:47
Ross Millikan
291k23196370
answered Nov 29 at 15:47
Ross Millikan
291k23196370
answered Nov 29 at 15:47
Ross Millikan
291k23196370
answered Nov 29 at 15:47
Ross Millikan
291k23196370
291k23196370
add a comment |
add a comment |
Another hint, assuming that your lg$(x)$ is $log_{a}(x)$. Note that $log_{a}(x)=ln(x)/ln(a)$...
answered Nov 29 at 15:54
Frobenius
613
Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
– mathishardd
Nov 29 at 16:00
1
Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
– Frobenius
Nov 29 at 16:05
add a comment |
Another hint, assuming that your lg$(x)$ is $log_{a}(x)$. Note that $log_{a}(x)=ln(x)/ln(a)$...
answered Nov 29 at 15:54
Frobenius
613
Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
– mathishardd
Nov 29 at 16:00
1
Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
– Frobenius
Nov 29 at 16:05
add a comment |
Another hint, assuming that your lg$(x)$ is $log_{a}(x)$. Note that $log_{a}(x)=ln(x)/ln(a)$...
answered Nov 29 at 15:54
Frobenius
613
Another hint, assuming that your lg$(x)$ is $log_{a}(x)$. Note that $log_{a}(x)=ln(x)/ln(a)$...
answered Nov 29 at 15:54
Frobenius
613
answered Nov 29 at 15:54
Frobenius
613
answered Nov 29 at 15:54
Frobenius
613
answered Nov 29 at 15:54
Frobenius
613
613
Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
– mathishardd
Nov 29 at 16:00
1
Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
– Frobenius
Nov 29 at 16:05
add a comment |
Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
– mathishardd
Nov 29 at 16:00
1
Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
– Frobenius
Nov 29 at 16:05
Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
– mathishardd
Nov 29 at 16:00
Thanks. Should I differentiate $ln((ln(x))/(ln(a)))$? or prove it some other way?
– mathishardd
Nov 29 at 16:00
1
1
Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
– Frobenius
Nov 29 at 16:05
Come on. Then $ln (ln x/ ln a)=ln ln x - ln ln a$. Don't you see the answer?
– Frobenius
Nov 29 at 16:05
add a comment |
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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 (!)
– user376343
Nov 29 at 15:36
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.
– Ross Millikan
Nov 29 at 15:45