Why isn't #log_b M/log_b N = log_b M - log_b N# true?

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Answer 1 · 2018-01-23T13:01:23

See below.

#log_bM/log_bN !=log_bM-log_bN#

#log_b(M/N)=log_bM-log_bN#

Reason:

If #p=log_a(x)# and #q=log_a(y)#

Then:

#a^p=x and a^q=y#

So:

#x/y=a^(p-q)#

i.e. #log_a(x/y)=p-q=>log_a(x/y)=log_a(x)-log_a(y)#

And:

#xy=a^(p+q)#

i.e. #log_a(xy)=p+q=>log_a(xy)=log_a(x)+log_a(y)#

#log_bM/log_bNcolor(white)(88)#

is the divisions of 2 logarithms, not the division of 2 numbers expressed to a common base.

And:

#log_bM-log_bN#

is the subtraction of 2 logarithms, which is how we divide 2 numbers expressed to a common base. i.e. by subtracting their powers.

To convince yourself:

#log_10(10)=1# and #log_10(100)=2#

#log_10(10)/log_10(100)=1/2#

#log_10(10)-log_10(100)=1-2=-1#

Hence:

#1/2!=-1#

So:

#log_10(10)/log_10(100)!=log_10(10)-log_10(100)#