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secondary 3 | A Maths
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logarithm
Date Posted: 4 years ago
Views: 284
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Here
Date Posted:
4 years ago
Sorry
Wrote wrongly for part a. Should be x > -2.
For part c, x cannot be 1 because the base for a logarithm cannot be 1 by definition, so 0 < x < 1 and 1 < x < 3 is correct.
In other words, there is no power for which 1 raised to that power equals 2.
Note, however, that 0 < x < 1 is alright because it is possible that numbers such as 0.5 can be raised to a negative power to yield 3 - 0.5.
Will update this on a new page tonight if no one has written a new copy yet.
Wrote wrongly for part a. Should be x > -2.
For part c, x cannot be 1 because the base for a logarithm cannot be 1 by definition, so 0 < x < 1 and 1 < x < 3 is correct.
In other words, there is no power for which 1 raised to that power equals 2.
Note, however, that 0 < x < 1 is alright because it is possible that numbers such as 0.5 can be raised to a negative power to yield 3 - 0.5.
Will update this on a new page tonight if no one has written a new copy yet.
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Hope this helps
Date Posted:
4 years ago
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