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secondary 4 | E Maths
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how do i solve this?
Date Posted: 4 years ago
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Date Posted:
4 years ago
thankyou!!
Do take note that the log in the equation refers to the natural logarithm (ln)
So the second last step should be :
x = e^log(32/3)
Then x = 32/3
So the second last step should be :
x = e^log(32/3)
Then x = 32/3
thankyou!
Ironically the "log" in Casio fx-96 SG Plus means "base 10".
Because it already uses ln for the natural log. It's the same for Sharp calculators also.
But in mathematical notation, if the base is not given, log refers to the natural log
But in mathematical notation, if the base is not given, log refers to the natural log
When the base of a logarithm is not explicitly written, it meant either (i) the context of the question made the interpretation unambiguous (for which we will have a specific base to use) or (ii) whichever base is unimportant. In this question, it is the latter. It is a matter of arbitrary choice between the common logarithm or the natural logarithm.
No matter how popular “log x” is used in Analysis to represent the natural logarithm of x, we are unable to say “log x” must refers to the natural logarithm. The ISO recommended notations of “lg” and “ln” would have made the “log x” specific but it may be the intent of the setter of the question to leave it as is for the student to decide and work around.
No matter how popular “log x” is used in Analysis to represent the natural logarithm of x, we are unable to say “log x” must refers to the natural logarithm. The ISO recommended notations of “lg” and “ln” would have made the “log x” specific but it may be the intent of the setter of the question to leave it as is for the student to decide and work around.
If the base is unimportant as you mentioned, then your second last step is unnecessary
It would simply be :
log x = log (32/3)
x = 32/3
Using either 10 or e (or any other base) would not be appropriate since the question has not specified (and given that you do not take it to specifically mean the natural log). We cannot simply choose the base (either 10 or e) we want and apply it, even though the result will be x = 32/3 for either.
The setter's intention would then be to test the student's understanding of additions and power manipulation of log terms with the same base (regardless of what the base is)
It would simply be :
log x = log (32/3)
x = 32/3
Using either 10 or e (or any other base) would not be appropriate since the question has not specified (and given that you do not take it to specifically mean the natural log). We cannot simply choose the base (either 10 or e) we want and apply it, even though the result will be x = 32/3 for either.
The setter's intention would then be to test the student's understanding of additions and power manipulation of log terms with the same base (regardless of what the base is)
@ Charmain, it is best to clarify with your teacher if the log in the question is meant to be ln, lg or meant to be arbitrary.
(I recommended e only because log x is most commonly taken to be ln x in mathematics. In the sciences (physics, chemistry, etc) and engineering, log x is usually to base 10 since it complements the usage of standard form(or scientific notation i.e x 10^n). And that is why the typical scientific calculator uses log for base 10 as well)
In your working, you can skip the second last step.
And for the current Singapore O level syllabus, (based on the question type you listed), you will always be given in the form ln or lg for these types of logarithms in the actual examination.
(I recommended e only because log x is most commonly taken to be ln x in mathematics. In the sciences (physics, chemistry, etc) and engineering, log x is usually to base 10 since it complements the usage of standard form(or scientific notation i.e x 10^n). And that is why the typical scientific calculator uses log for base 10 as well)
In your working, you can skip the second last step.
And for the current Singapore O level syllabus, (based on the question type you listed), you will always be given in the form ln or lg for these types of logarithms in the actual examination.
alright, thankyou so much!
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