Calculate the factor increase for y=ln x
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Question:
Suppose $x >1$ and $x$ increases by a factor of $5$. By what factor will $y$ increase given $y = ln(x)$?
My answer:
So the increase would be $$frac{(ln(5x) - ln x)}{ln x} times 100%$$
And then $$frac{( ln 5 + ln x -ln x )}{ln x}times 100 %$$
But thats clearly wrong because $ln x$ get cancelled in the numerator but the denominator is $ln x$
logarithms
edited Dec 13 '18 at 19:33
Emilio Novati
52.1k43474
asked Dec 13 '18 at 17:22
LJacobLJacob
61
$endgroup$
add a comment |
$begingroup$
Question:
Suppose $x >1$ and $x$ increases by a factor of $5$. By what factor will $y$ increase given $y = ln(x)$?
My answer:
So the increase would be $$frac{(ln(5x) - ln x)}{ln x} times 100%$$
And then $$frac{( ln 5 + ln x -ln x )}{ln x}times 100 %$$
But thats clearly wrong because $ln x$ get cancelled in the numerator but the denominator is $ln x$
logarithms
edited Dec 13 '18 at 19:33
Emilio Novati
52.1k43474
asked Dec 13 '18 at 17:22
LJacobLJacob
61
$endgroup$
$begingroup$
Why must that be wrong? The factor of increase in $y$ will not be independent of $x$. Try some numerical examples. (Please edit the question to reformat with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:26
add a comment |
$begingroup$
Question:
Suppose $x >1$ and $x$ increases by a factor of $5$. By what factor will $y$ increase given $y = ln(x)$?
My answer:
So the increase would be $$frac{(ln(5x) - ln x)}{ln x} times 100%$$
And then $$frac{( ln 5 + ln x -ln x )}{ln x}times 100 %$$
But thats clearly wrong because $ln x$ get cancelled in the numerator but the denominator is $ln x$
logarithms
edited Dec 13 '18 at 19:33
Emilio Novati
52.1k43474
asked Dec 13 '18 at 17:22
LJacobLJacob
61
$endgroup$
Question:
Suppose $x >1$ and $x$ increases by a factor of $5$. By what factor will $y$ increase given $y = ln(x)$?
My answer:
So the increase would be $$frac{(ln(5x) - ln x)}{ln x} times 100%$$
And then $$frac{( ln 5 + ln x -ln x )}{ln x}times 100 %$$
But thats clearly wrong because $ln x$ get cancelled in the numerator but the denominator is $ln x$
logarithms
logarithms
edited Dec 13 '18 at 19:33
Emilio Novati
52.1k43474
asked Dec 13 '18 at 17:22
LJacobLJacob
61
edited Dec 13 '18 at 19:33
Emilio Novati
52.1k43474
asked Dec 13 '18 at 17:22
LJacobLJacob
61
edited Dec 13 '18 at 19:33
Emilio Novati
52.1k43474
edited Dec 13 '18 at 19:33
Emilio Novati
52.1k43474
edited Dec 13 '18 at 19:33
Emilio Novati
52.1k43474
52.1k43474
asked Dec 13 '18 at 17:22
LJacobLJacob
61
asked Dec 13 '18 at 17:22
LJacobLJacob
61
asked Dec 13 '18 at 17:22
LJacobLJacob
61
61
$begingroup$
Why must that be wrong? The factor of increase in $y$ will not be independent of $x$. Try some numerical examples. (Please edit the question to reformat with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:26
add a comment |
$begingroup$
Why must that be wrong? The factor of increase in $y$ will not be independent of $x$. Try some numerical examples. (Please edit the question to reformat with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:26
$begingroup$
Why must that be wrong? The factor of increase in $y$ will not be independent of $x$. Try some numerical examples. (Please edit the question to reformat with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:26
$begingroup$
Why must that be wrong? The factor of increase in $y$ will not be independent of $x$. Try some numerical examples. (Please edit the question to reformat with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:26
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The increase in $y$, is just given by $ Delta y=ln(5x)-ln x=ln 5$, which is independent of the value of $x$.
The relative increase in $y$ is given by $frac{Delta y}y=ln5/ln x$, which is not independent of $x$, and can also be expressed as a percentage.
answered Dec 13 '18 at 17:31
Shubham JohriShubham Johri
5,204718
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The increase in $y$, is just given by $ Delta y=ln(5x)-ln x=ln 5$, which is independent of the value of $x$.
The relative increase in $y$ is given by $frac{Delta y}y=ln5/ln x$, which is not independent of $x$, and can also be expressed as a percentage.
answered Dec 13 '18 at 17:31
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
$begingroup$
The increase in $y$, is just given by $ Delta y=ln(5x)-ln x=ln 5$, which is independent of the value of $x$.
The relative increase in $y$ is given by $frac{Delta y}y=ln5/ln x$, which is not independent of $x$, and can also be expressed as a percentage.
answered Dec 13 '18 at 17:31
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
$begingroup$
The increase in $y$, is just given by $ Delta y=ln(5x)-ln x=ln 5$, which is independent of the value of $x$.
The relative increase in $y$ is given by $frac{Delta y}y=ln5/ln x$, which is not independent of $x$, and can also be expressed as a percentage.
answered Dec 13 '18 at 17:31
Shubham JohriShubham Johri
5,204718
$endgroup$
The increase in $y$, is just given by $ Delta y=ln(5x)-ln x=ln 5$, which is independent of the value of $x$.
The relative increase in $y$ is given by $frac{Delta y}y=ln5/ln x$, which is not independent of $x$, and can also be expressed as a percentage.
answered Dec 13 '18 at 17:31
Shubham JohriShubham Johri
5,204718
answered Dec 13 '18 at 17:31
Shubham JohriShubham Johri
5,204718
answered Dec 13 '18 at 17:31
Shubham JohriShubham Johri
5,204718
answered Dec 13 '18 at 17:31
Shubham JohriShubham Johri
5,204718
5,204718
add a comment |
add a comment |
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$begingroup$
Why must that be wrong? The factor of increase in $y$ will not be independent of $x$. Try some numerical examples. (Please edit the question to reformat with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 13 '18 at 17:26