slope of a linear function in semi-log plot
$begingroup$
I have a decreasing linear function. So,the slope of this function df/dt <0. Now, if we plot this function in a semi-log plot with log(t) in horizontal axis and y in vertical axis, can we say that this is still a decreasing function with df/d(logt)<0?
logarithms graphing-functions slope
edited Jan 13 at 15:04
daw
25.1k1745
asked Jan 13 at 0:19
Gate oxideGate oxide
112
$endgroup$
add a comment |
$begingroup$
I have a decreasing linear function. So,the slope of this function df/dt <0. Now, if we plot this function in a semi-log plot with log(t) in horizontal axis and y in vertical axis, can we say that this is still a decreasing function with df/d(logt)<0?
logarithms graphing-functions slope
edited Jan 13 at 15:04
daw
25.1k1745
asked Jan 13 at 0:19
Gate oxideGate oxide
112
$endgroup$
add a comment |
$begingroup$
I have a decreasing linear function. So,the slope of this function df/dt <0. Now, if we plot this function in a semi-log plot with log(t) in horizontal axis and y in vertical axis, can we say that this is still a decreasing function with df/d(logt)<0?
logarithms graphing-functions slope
edited Jan 13 at 15:04
daw
25.1k1745
asked Jan 13 at 0:19
Gate oxideGate oxide
112
$endgroup$
I have a decreasing linear function. So,the slope of this function df/dt <0. Now, if we plot this function in a semi-log plot with log(t) in horizontal axis and y in vertical axis, can we say that this is still a decreasing function with df/d(logt)<0?
logarithms graphing-functions slope
logarithms graphing-functions slope
edited Jan 13 at 15:04
daw
25.1k1745
asked Jan 13 at 0:19
Gate oxideGate oxide
112
edited Jan 13 at 15:04
daw
25.1k1745
asked Jan 13 at 0:19
Gate oxideGate oxide
112
edited Jan 13 at 15:04
daw
25.1k1745
edited Jan 13 at 15:04
daw
25.1k1745
edited Jan 13 at 15:04
daw
25.1k1745
25.1k1745
asked Jan 13 at 0:19
Gate oxideGate oxide
112
asked Jan 13 at 0:19
Gate oxideGate oxide
112
asked Jan 13 at 0:19
Gate oxideGate oxide
112
112
add a comment |
add a comment |
2 Answers
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$begingroup$
Yes. $sle tRightarrowlog(s)le log(t)Rightarrow f(log(s)) ge f(log(t))$, since $log$ is monotonically increasing and $f$ is monotonically decreasing.
answered Jan 13 at 0:25
user408858user408858
479213
$endgroup$
add a comment |
$begingroup$
Consider
$$frac{dy}{dt}=frac{dy}{dlog(t)}times frac{dlog(t)}{dt}=frac 1tfrac{dy}{dlog(t)}$$
$$frac{dy}{dlog(t)}=tfrac{dy}{dt}$$ Then $cdots$
answered Jan 13 at 3:14
Claude LeiboviciClaude Leibovici
126k1158135
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Yes. $sle tRightarrowlog(s)le log(t)Rightarrow f(log(s)) ge f(log(t))$, since $log$ is monotonically increasing and $f$ is monotonically decreasing.
answered Jan 13 at 0:25
user408858user408858
479213
$endgroup$
add a comment |
$begingroup$
Yes. $sle tRightarrowlog(s)le log(t)Rightarrow f(log(s)) ge f(log(t))$, since $log$ is monotonically increasing and $f$ is monotonically decreasing.
answered Jan 13 at 0:25
user408858user408858
479213
$endgroup$
add a comment |
$begingroup$
Yes. $sle tRightarrowlog(s)le log(t)Rightarrow f(log(s)) ge f(log(t))$, since $log$ is monotonically increasing and $f$ is monotonically decreasing.
answered Jan 13 at 0:25
user408858user408858
479213
$endgroup$
Yes. $sle tRightarrowlog(s)le log(t)Rightarrow f(log(s)) ge f(log(t))$, since $log$ is monotonically increasing and $f$ is monotonically decreasing.
answered Jan 13 at 0:25
user408858user408858
479213
answered Jan 13 at 0:25
user408858user408858
479213
answered Jan 13 at 0:25
user408858user408858
479213
answered Jan 13 at 0:25
user408858user408858
479213
479213
add a comment |
add a comment |
$begingroup$
Consider
$$frac{dy}{dt}=frac{dy}{dlog(t)}times frac{dlog(t)}{dt}=frac 1tfrac{dy}{dlog(t)}$$
$$frac{dy}{dlog(t)}=tfrac{dy}{dt}$$ Then $cdots$
answered Jan 13 at 3:14
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
add a comment |
$begingroup$
Consider
$$frac{dy}{dt}=frac{dy}{dlog(t)}times frac{dlog(t)}{dt}=frac 1tfrac{dy}{dlog(t)}$$
$$frac{dy}{dlog(t)}=tfrac{dy}{dt}$$ Then $cdots$
answered Jan 13 at 3:14
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
add a comment |
$begingroup$
Consider
$$frac{dy}{dt}=frac{dy}{dlog(t)}times frac{dlog(t)}{dt}=frac 1tfrac{dy}{dlog(t)}$$
$$frac{dy}{dlog(t)}=tfrac{dy}{dt}$$ Then $cdots$
answered Jan 13 at 3:14
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
Consider
$$frac{dy}{dt}=frac{dy}{dlog(t)}times frac{dlog(t)}{dt}=frac 1tfrac{dy}{dlog(t)}$$
$$frac{dy}{dlog(t)}=tfrac{dy}{dt}$$ Then $cdots$
answered Jan 13 at 3:14
Claude LeiboviciClaude Leibovici
126k1158135
answered Jan 13 at 3:14
Claude LeiboviciClaude Leibovici
126k1158135
answered Jan 13 at 3:14
Claude LeiboviciClaude Leibovici
126k1158135
answered Jan 13 at 3:14
Claude LeiboviciClaude Leibovici
126k1158135
126k1158135
add a comment |
add a comment |
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