How do you solve \[\log \left( {\dfrac{1}{{100}}} \right) = \log ({10^{x + 2}})\] ?

Hint: We can solve this using the rules or laws of logarithms. If we have \[\log (x) = \log (y)\] we can cancel the logarithmic function we get \[ \Rightarrow x = y\] . We have the logarithm rule \[\log {x^a} = a\log x\] . This is also called power rule. The logarithm of an exponential number is the exponent times the logarithm of the base. Using this we can solve this. We need to find the value of ‘x’.

Complete step-by-step answer:
We have \[\log \left( {\dfrac{1}{{100}}} \right) = \log ({10^{x + 2}})\] .
We know that if \[\log (x) = \log (y)\] then \[x = y\] . Comparing this we have \[x = \dfrac{1}{{100}}\] and \[y = {10^{x + 2}}\] .
Then we have,
 \[ \Rightarrow \dfrac{1}{{100}} = {10^{x + 2}}\]
 \[ \Rightarrow \dfrac{1}{{{{10}^2}}} = {10^{x + 2}}\]
It can be written as
 \[ \Rightarrow {10^{ - 2}} = {10^{x + 2}}\]
Applying logarithm on both sides we have,
 \[ \Rightarrow \log ({10^{ - 2}}) = \log ({10^{x + 2}})\]
We know \[\log {x^a} = a\log x\] , then above becomes:
 \[ \Rightarrow - 2\log (10) = (x + 2)\log (10)\]
We know that the value of \[\log (10)\] is 1.
 \[ \Rightarrow - 2 = (x + 2)\]
Rearranging the above equation we have,
 \[ \Rightarrow x + 2 = - 2\]
Subtracting 2 on both sides we have,
 \[ \Rightarrow x = - 2 - 2\]
 \[ \Rightarrow x = - 4\] . Is the required answer.
(Here we note that the base is 10.)
So, the correct answer is “ x = - 4”.

Note: To solve this kind of problem we need to remember the laws of logarithms. Product rule of logarithm that is the logarithm of the product is the sum of the logarithms of the factors. That is \[\log (x.y) = \log (x) + \log (y)\] . Quotient rule of logarithm that is the logarithm of the ratio of two quantities is the logarithm of the numerator minus the logarithm of the denominator. that is \[\log \left( {\dfrac{x}{y}} \right) = \log x - \log y\] . Power rule of logarithm that is the logarithm of an exponential number is the exponent times the logarithm of the base. That is \[\log {x^a} = a\log x\] . These are the basic rules we use while solving a problem that involves logarithm function.