Question
How would you solve this equation?
log(ab)=log(a)*log(b)
a=30, b=60
Does the it appear that this is a true statement? Why?
log(ab)=log(a)*log(b)
a=30, b=60
Does the it appear that this is a true statement? Why?
Answers
bobpursley
The equation is not true. Consider a=10, b=1
log(10*1)= log10 *log1
and since the log of 1 is zero, this makes the log of 10 equal to zero, so it is nonsense.
Try log(ab)=log a+ log b
log(10*1)= log10 *log1
and since the log of 1 is zero, this makes the log of 10 equal to zero, so it is nonsense.
Try log(ab)=log a+ log b
Reiny
other than the obvious of
a=1, b=1, i found some that come very close.
I was so intrigued by the question that I actually went ahead and made up a short computer program to test this.
This is in a very archaic language called "BASIC", but I am sure some of the tutors will recognize it
FOR A = .01 TO 100 STEP .01
FOR B = .01 TO 100 STEP .01
LS = LOG(A*B)/LOG(10)
RS = LOG(A)*LOG(B)/(LOG(10)*LOG(10)
IF ABS(LS-RS) < .00001 THEN PRINT A,B
NEXT B
NEXT A
I got a = .28 and b= 2.27 to have an error of less than .00001 between the left side and the right side of your equation.
of course a=2.27 and b=.28 will also work within that margin of error.
a=1, b=1, i found some that come very close.
I was so intrigued by the question that I actually went ahead and made up a short computer program to test this.
This is in a very archaic language called "BASIC", but I am sure some of the tutors will recognize it
FOR A = .01 TO 100 STEP .01
FOR B = .01 TO 100 STEP .01
LS = LOG(A*B)/LOG(10)
RS = LOG(A)*LOG(B)/(LOG(10)*LOG(10)
IF ABS(LS-RS) < .00001 THEN PRINT A,B
NEXT B
NEXT A
I got a = .28 and b= 2.27 to have an error of less than .00001 between the left side and the right side of your equation.
of course a=2.27 and b=.28 will also work within that margin of error.