How do you evaluate #Log_10 (1/10^x)#?

2 Answers
Jan 25, 2016

Answer:

I found: #-x#

Explanation:

You can use the property of logs that says:
#logx-logy=log(x/y)#
and write:

#log_(10)(1/10^x)=log_(10)1-log_(10)10^x=#

we then use the definition of log:

#log_ax=y ->x=a^y#

and solve each of them:
#log_(10)1=0#
#log_(10)10^x=x#
so:
#log_(10)(1/10^x)=0-x=-x#

Jan 25, 2016

Answer:

# log_10(1 / 10^x) = -x #

Explanation:

First of all, use the logarithmic law

#log_a x - log_a y = log_a (x/y) #

Thus,

#log_10(1 / 10^x) = log_10(1) - log_10(10^x)#

Now,

#log_a(1) = 0#

holds for any base #a#, so you can omit #log_10(1)# and have:

#log_10(1 / 10^x) = log_10(1) - log_10(10^x) = - log_10(10^x)#

As last, #log_a(x)# and #a^x# are inverse functions for any base #a != 1#, #a > 0#.

What does this mean exactly? It means that both #log_a(a^r) = x# and #a^(log_a(x)) = x# hold for any positive number #x#.

Basically, if you take any positive real number #x# and first exponentiate it and then take the logarithm of the same basis, the two operations "exponentiate" and "taking the logarithm" eliminate each other and in the end, you have just #x# left.

The same holds if you first take the logarithm and then exponentiate. In this case, the two operations also "cancel" each other and you also end up with your original number #x#.

Example:

Say #x = 8# and the basis is #2#.

If you first compute #2^x = 2^8 = 256# and later compute #log_2(2^x)#, you will have
#log_2(2^x) = log_2(256) = 8 = x#

The same thing happens if you first compute #log_2(x) = log_2(8) = 3# and then afterwards, exponentiate #2^(log_2(x)) = 2^3 = 8 = x#

So, let's get back to your question:

#log_10(x)# and #10^x# are inverse functions which means that

#log_10(10^x) = x#.

So, in total, you have:

#log_10(1 / 10^x) = log_10(1) - log_10(10^x) = - log_10(10^x) = -x#