How do you calculate # log 0.1 #?

1 Answer

Answer:

#log_(10)(0.1)=-1# - or in other words, we take the 10 and flip flop it to the denominator of a fraction where we have #1/10#.

Explanation:

Let's think about this question in a different way than it's being asked - I find that sometimes students understand exponents and powers better than they understand logs.

The term #log(0.1)# is short for #log_(10)(0.1)# and asks the question - how many times do I need to multiply 10 by itself in order to get to #0.1#. A different way of seeing this same question is by asking this:

#10^x=0.1#

So the above and

#log_(10)(0.1)#

are the same question - it's just that in that first one we need to solve for #x# and the second one is a statement of a value.

So what do they equal?

Let's solve the exponent question first and then the statement of value will become clear:

#10^x=0.1=1/10#

At this point it'd be helpful to know that when we have a negative exponent, it means that we're talking about a fractional value and that the value that has the fractional exponent, to be positive, needs to swap its place in the fraction (so move to the denominator from the numerator, or vice versa).

So the expression #10^-1# means that this term is in a fraction and for the exponent to be positive, the term needs to swap places. Like this:

#10^-1=10^-1/1=1/10^1=1/10#

So #x=-1#. And that is the answer to the statement of value - the log term:

#log_(10)(0.1)=-1# - or in other words, we take the 10 and flip flop it to the denominator of a fraction where we have #1/10#.