Maths Question

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Nithu05
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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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(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
If  y=e^x then you're absolutely right that  ln(y)=x

But as you also say, by definition  ln(x)=log_e (x)

Is the root of your confusion an ambiguous mark scheme or something like that?

Or perhaps its just that you don't see that  y=e^x 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)  ln(x)=log_e (x)
Last edited by jon nicholls; 1 month ago
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Nithu05
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(Original post by jon nicholls)
If  y=e^x then you're absolutely right that  ln(y)=x

But as you also say, by definition  ln(x)=log_e (x)

Is the root of your confusion an ambiguous mark scheme or something like that?

Or perhaps its just that you don't see that  y=e^x 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)  ln(x)=log_e (x)
Thank you so much. It was just the way it was worded in my textbook that made me think there was a link between the equation and the natural log. Thanks for taking the time to explain
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Nithu05
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(Original post by jon nicholls)
If  y=e^x then you're absolutely right that  ln(y)=x

But as you also say, by definition  ln(x)=log_e (x)

Is the root of your confusion an ambiguous mark scheme or something like that?

Or perhaps its just that you don't see that  y=e^x 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)  ln(x)=log_e (x)
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
Last edited by Nithu05; 1 month ago
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jon nicholls
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(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
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:  ln(\frac{1}{e})=-1

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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(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:  ln(\frac{1}{e})=-1

But log(x) is undefined if x<0 *once again think of the graph*.

Hope that helps.
Thank you so much This really helped
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Nithu05
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(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:  ln(\frac{1}{e})=-1

But log(x) is undefined if x<0 *once again think of the graph*.

Hope that helps.
So x cannot be -2 right in the picture attached?
Attached files
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davros
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(Original post by Nithu05)
So x cannot be -2 right in the picture attached?
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
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jon nicholls
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'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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Nithu05
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(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
(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/
Thank you so much for the help guys
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