How do you solve \[{\log _{10}}4 + {\log _{10}}25\] ?

Hint: When one term is raised to the power of another term, the function is called an exponential function, for example \[a = {x^y}\] . The inverse of the exponential functions are called logarithm functions, the inverse of the function given in the example is \[y = {\log _x}a\] that is a logarithm function. These functions also obey certain rules called laws of the logarithm, using these laws we can write the function in a variety of ways. Using the laws of the logarithm, we can solve the given problem.

Complete step-by-step answer:
Given,
 \[{\log _{10}}4 + {\log _{10}}25\]
Since the base is the same, we can apply the logarithm laws.
We know the product of logarithm, that is \[\log (x.y) = \log (x) + \log (y)\] .
where \[x = 4\] and \[y = 25\] . Substituting we have,
 \[{\log _{10}}4 + {\log _{10}}25 = {\log _{10}}(4 \times 25)\]
 \[ = {\log _{10}}(100)\]
 \[ = {\log _{10}}({10^2})\]
Again we know the power rule of logarithm, that is \[\log {x^a} = a\log x\] . applying we have,
 \[ = 2{\log _{10}}(10)\]
We know \[{\log _{10}}(10) = 1\]
 \[ = 2\]
Thus we have, \[{\log _{10}}4 + {\log _{10}}25 = 2\]
So, the correct answer is “ 2”.

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.