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

##### 2 Answers

I found:

#### Explanation:

You can use the property of logs that says:

and write:

we then use the definition of log:

and solve each of them:

so:

#### 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

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

As last,

What does this mean exactly? It means that both

Basically, if you take any positive real number

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

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(10^x) = x# .

So, in total, you have: