Question

Simplify: \[{{\log }_{9}}729+{{\log }_{9}}81+{{\log }_{9}}9\]

Answer 1

Hint: To solve these kinds of problems, we need to know about the basic formulas related to logarithm. Firstly, we assume the given expression as ‘E’. This problem is solved step by step using related formulas and identities and thus obtaining the answer. Also, it is very important for us to know the properties of logarithm and which we are used in this problem:
\[{{\log }_{a}}{{x}^{n}}=n{{\log }_{a}}x\]
\[{{\log }_{a}}1=0\]
\[{{\log }_{a}}{{a}^{r}}=r\]
\[{{\log }_{a}}a=1\]

Complete step by step answer:
Logarithm is the exponent that indicates the power to which a base number is raised to produce a given number Or A logarithmic is defined as the power to which number which must be raised to get some values. Logarithmic functions and exponential functions are inverses to each other. We should be clear when the rule is applied to the power, then the exponent rule is used.
Example: The logarithm of 100 to the base 10 is 2.
As given in the question,
\[\Rightarrow {{\log }_{9}}729+{{\log }_{9}}81+{{\log }_{9}}9\]
As we know that, \[729={{9}^{3}}\]and \[81={{9}^{2}}\], we get the above expression as
 \[\Rightarrow {{\log }_{9}}\left( {{9}^{3}} \right)+{{\log }_{9}}\left( {{9}^{2}} \right)+{{\log }_{9}}9\]
According to logarithmic property, \[{{\log }_{a}}{{x}^{n}}=n{{\log }_{a}}x\] , we get
\[\Rightarrow 3{{\log }_{9}}9+2{{\log }_{9}}9+{{\log }_{9}}9\]
For further evaluation,
We are using the logarithm identity \[{{\log }_{a}}a=1\], we get
\[\Rightarrow 3+2+1\]
\[=6\].
Therefore, \[{{\log }_{9}}729+{{\log }_{9}}81+{{\log }_{9}}9=6\].

Note: One must remember the log values of arguments from 1 to 10 with the base value of the log equal to 10. More generally, exponentiation allow any positive real number as base to be raised to any real power, always producing a positive result, so \[{{\log }_{b}}x\] for any two positive real numbers b and x where b is not equal to 1, is always a unique real number y.