Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
secondary 3 | A Maths
One Answer Below
Anyone can contribute an answer, even non-tutors.
need help with this qn, got stucked because log 2 9 cannot be simplified (i did change of base to log 4 9)
Date Posted: 3 years ago
Views: 506
See 1 Answer
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Interestingly, had the question started as log2 x^2 and not 2 log2 x, the negative solution of x would be valid.
Date Posted:
3 years ago
LockB, for such questions we don’t usually compute values of things like log2 9. For these kind of “solve for x” questions, we tend to combine logs together rather than evaluating logs directly unless you reach an equation like e^x = 5 which you try to solve by logs and putting in calculator.
btw for the defined and undefined thing, how do we know whether its defined or not, i always thought if the answer is negative then it would be considered as undefined
log a b = c
a must be a positive number not equaling 1
b must be a positive number
c is unrestricted
So just now we obtained
log2 x = number
This is on the other side of the log, so it’s still possible to be negative and have solution
a must be a positive number not equaling 1
b must be a positive number
c is unrestricted
So just now we obtained
log2 x = number
This is on the other side of the log, so it’s still possible to be negative and have solution
The definition of log and the conditions are all given in the textbook. You'll have to do some revision.
my textbook only said that in log a q, q have to be more than 0
Depends on the book.
The sad thing about negative bases is that they can only be raised to integer powers. I can’t imagine a negative power being raised to a “decimal”. Likely this falls into complex numbers.
Usually we only consider positive bases for our syllabus.
Probably nowhere in the book did any log to a negative base is written down at all.
The sad thing about negative bases is that they can only be raised to integer powers. I can’t imagine a negative power being raised to a “decimal”. Likely this falls into complex numbers.
Usually we only consider positive bases for our syllabus.
Probably nowhere in the book did any log to a negative base is written down at all.
Yes, and that info in your textbook is already sufficient for your level. Because they will always give you a base that is above 1 for O levels.
Actually it can still work as long even if a and b are negative, as long as b is an exponent of a.
log a b = c comes from b = a^c
If b = -125, and -125 = (-5)³
Then a = -5 and c = 3
log a b = log(-5) (-5)³
= 3 log(-5) (-5)
= 3
But whether this is actually Acceptable remains to be seen
Actually it can still work as long even if a and b are negative, as long as b is an exponent of a.
log a b = c comes from b = a^c
If b = -125, and -125 = (-5)³
Then a = -5 and c = 3
log a b = log(-5) (-5)³
= 3 log(-5) (-5)
= 3
But whether this is actually Acceptable remains to be seen
thx :)
https://math.stackexchange.com/questions/690024/why-must-the-base-of-a-logarithm-be-a-positive-real-number-not-equal-to-1 the answer is probably found here...
In the older textbooks, there would be the explanation of why the base cannot be 1 :
Because the logarithm is the inverse function of the exponential operation
If a^b = c, then loga c = b
if a = 1, then 1^b = c and log1 c = b
But 1^b always = 1 for any real value of b.
This also means c can only = 1.
Then, b can be any number so this means that log1 (1) can be any real number. So this is not specific
In the older textbooks, there would be the explanation of why the base cannot be 1 :
Because the logarithm is the inverse function of the exponential operation
If a^b = c, then loga c = b
if a = 1, then 1^b = c and log1 c = b
But 1^b always = 1 for any real value of b.
This also means c can only = 1.
Then, b can be any number so this means that log1 (1) can be any real number. So this is not specific
360 Tutoring Program