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Nithu05
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#1
Hey guys, I'm currently studying natural logs and I am confused as to why the lnx= log_e(x) rather than log_e(y) beacuse y=e^x. Could you please explain? Thank you so much
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jon nicholls
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#2
(Original post by Nithu05)
Hey guys, I'm currently studying natural logs and I am confused as to why the lnx= log_e(x) rather than log_e(y) beacuse y=e^x. Could you please explain? Thank you so much
Hey guys, I'm currently studying natural logs and I am confused as to why the lnx= log_e(x) rather than log_e(y) beacuse y=e^x. Could you please explain? Thank you so much
then you're absolutely right that 
But as you also say, by definition
Is the root of your confusion an ambiguous mark scheme or something like that?
Or perhaps its just that you don't see that
is not always the case. The question would have to give you this connection between the two variables x and y. But by definition, (so its always the case)
Last edited by jon nicholls; 1 month ago
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Nithu05
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#3
(Original post by jon nicholls)
If then you're absolutely right that
But as you also say, by definition
Is the root of your confusion an ambiguous mark scheme or something like that?
Or perhaps its just that you don't see that is not always the case. The question would have to give you this connection between the two variables x and y. But by definition, (so its always the case)
If
then you're absolutely right that But as you also say, by definition
Is the root of your confusion an ambiguous mark scheme or something like that?
Or perhaps its just that you don't see that
is not always the case. The question would have to give you this connection between the two variables x and y. But by definition, (so its always the case)
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Nithu05
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#4
(Original post by jon nicholls)
If then you're absolutely right that
But as you also say, by definition
Is the root of your confusion an ambiguous mark scheme or something like that?
Or perhaps its just that you don't see that is not always the case. The question would have to give you this connection between the two variables x and y. But by definition, (so its always the case)
If
then you're absolutely right that But as you also say, by definition
Is the root of your confusion an ambiguous mark scheme or something like that?
Or perhaps its just that you don't see that
is not always the case. The question would have to give you this connection between the two variables x and y. But by definition, (so its always the case)
Last edited by Nithu05; 1 month ago
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jon nicholls
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#5
(Original post by Nithu05)
One more question, when would the solution be negative? Can x only take that value if it’s a natural log? Edit: Wrong Post sorry
One more question, when would the solution be negative? Can x only take that value if it’s a natural log? Edit: Wrong Post sorry
Here's the cheat sheet:
log(x)<0 when 0<x<1 *think of the graph of y=log(x)*
Example:
But log(x) is undefined if x<0 *once again think of the graph*.
Hope that helps.
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Nithu05
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#6
(Original post by jon nicholls)
So you can obtain negative values when inverting a log using exponentiation, but you can't 'log' a negative number (because the curve will not be continuous but that's a long complicated story).
Here's the cheat sheet:
log(x)<0 when 0<x<1 *think of the graph of y=log(x)*
Example:
But log(x) is undefined if x<0 *once again think of the graph*.
Hope that helps.
So you can obtain negative values when inverting a log using exponentiation, but you can't 'log' a negative number (because the curve will not be continuous but that's a long complicated story).
Here's the cheat sheet:
log(x)<0 when 0<x<1 *think of the graph of y=log(x)*
Example:
But log(x) is undefined if x<0 *once again think of the graph*.
Hope that helps.
This really helped
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Nithu05
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#7
(Original post by jon nicholls)
So you can obtain negative values when inverting a log using exponentiation, but you can't 'log' a negative number (because the curve will not be continuous but that's a long complicated story).
Here's the cheat sheet:
log(x)<0 when 0<x<1 *think of the graph of y=log(x)*
Example:
But log(x) is undefined if x<0 *once again think of the graph*.
Hope that helps.
So you can obtain negative values when inverting a log using exponentiation, but you can't 'log' a negative number (because the curve will not be continuous but that's a long complicated story).
Here's the cheat sheet:
log(x)<0 when 0<x<1 *think of the graph of y=log(x)*
Example:
But log(x) is undefined if x<0 *once again think of the graph*.
Hope that helps.
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davros
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#8
(Original post by Nithu05)
So x cannot be -2 right in the picture attached?
So x cannot be -2 right in the picture attached?
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jon nicholls
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#9
'Extraneous' solutions like Davros is describing are tricky to spot and demonstrate why you should always 'test' your solutions by plugging them back into the original equation to identify any which may be incorrect. They arise for a number of separate subtle reasons.
https://www.themathdoctors.org/extra...ses-and-cures/
https://www.themathdoctors.org/extra...ses-and-cures/
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Nithu05
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#10
(Original post by davros)
This is answered in your other thread for that question - when you convert 2log(x) to log(x^2) you are basically introducing a second, negative solution which doesn't satisfy the original equation
This is answered in your other thread for that question - when you convert 2log(x) to log(x^2) you are basically introducing a second, negative solution which doesn't satisfy the original equation
(Original post by jon nicholls)
'Extraneous' solutions like Davros is describing are tricky to spot and demonstrate why you should always 'test' your solutions by plugging them back into the original equation to identify any which may be incorrect. They arise for a number of separate subtle reasons.
https://www.themathdoctors.org/extra...ses-and-cures/
'Extraneous' solutions like Davros is describing are tricky to spot and demonstrate why you should always 'test' your solutions by plugging them back into the original equation to identify any which may be incorrect. They arise for a number of separate subtle reasons.
https://www.themathdoctors.org/extra...ses-and-cures/
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