Question

Simplify: \[{\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}}\]

Answer 1

Hint:
Here, we have to simplify the given logarithmic expression. We will use the rules of the logarithmic and exponent to simplify the expression. An expression is a sentence with the minimum of two numbers or more. These numbers are related using an arithmetic operation like addition, subtraction, multiplication or division.

Formula Used:
We will use the following formulas:
1) Exponent rule 1: \[{\left( {{a^m}} \right)^n} = {a^{mn}}\]
2) Exponent rule 2: \[{a^m} \cdot {a^n} = {a^{m + n}}\]
3) Exponent rule 3: \[{a^{ - m}} = \dfrac{1}{{{a^m}}}\]
4) Logarithmic rule 1: \[{\log _{\dfrac{1}{a}}}a = - {\log _a}a\]
5) Logarithmic rule 2: \[{\log _a}a = - 1\]

Complete step by step solution:
We are given a function \[{\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}}\]
By rewriting the given expression, we get
\[{\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}} = {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{{\left( {{3^2}} \right)}^{ - 1}}{{\left( {{3^3}} \right)}^{\dfrac{{ - 4}}{3}}}}}}}\]
By using the exponent rule \[{\left( {{a^m}} \right)^n} = {a^{mn}}\], we get
\[ \Rightarrow {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}} = {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{{\left( 3 \right)}^{ - 2}}{{\left( 3 \right)}^{\dfrac{{ - 12}}{3}}}}}}}\]
By simplifying the exponents, we get
\[ \Rightarrow {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}} = {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{{\left( 3 \right)}^{ - 2}}{{\left( 3 \right)}^{ - 4}}}}}}\]
By using the exponent rule \[{a^m} \cdot {a^n} = {a^{m + n}}\], we get
\[ \Rightarrow {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}} = {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{{\left( 3 \right)}^{ - 2 - 4}}}}}}\]
Adding the exponents with the same base, we get
\[ \Rightarrow {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}} = {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{{\left( 3 \right)}^{ - 6}}}}}}\]
Now using the exponent rule \[{\left( {{a^m}} \right)^n} = {a^{mn}}\], we get
\[ \Rightarrow {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}} = {\log _{\dfrac{1}{3}}}\sqrt[4]{{729 \times {{\left( 3 \right)}^{ - 6 \times \dfrac{1}{3}}}}}\]
By dividing the exponents, we get
\[ \Rightarrow {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}} = {\log _{\dfrac{1}{3}}}\sqrt[4]{{729 \times {{\left( 3 \right)}^{ - 2}}}}\]
Rewriting the equation, we get
\[ \Rightarrow {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}} = {\log _{\dfrac{1}{3}}}\sqrt[4]{{729 \times \dfrac{1}{9}}}\]
By dividing the term, we get
\[ \Rightarrow {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}} = {\log _{\dfrac{1}{3}}}\sqrt[4]{{81}}\]
Again rewriting in the form of exponents, we get
\[ \Rightarrow {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}} = {\log _{\dfrac{1}{3}}}\sqrt[4]{{{3^4}}}\]
By rewriting the surds in the form of exponents, we get
\[ \Rightarrow {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}} = {\log _{\dfrac{1}{3}}}{\left( {{3^4}} \right)^{\dfrac{1}{4}}}\]
By using the exponent rule \[{\left( {{a^m}} \right)^n} = {a^{mn}}\], we get
\[ \Rightarrow {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}} = {\log _{\dfrac{1}{3}}}\left( {{3^{4 \times \dfrac{1}{4}}}} \right)\]
By dividing the exponents, we get
\[ \Rightarrow {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}} = {\log _{\dfrac{1}{3}}}\left( 3 \right)\]
By using the exponent rule \[{a^{ - m}} = \dfrac{1}{{{a^m}}}\], we get
\[ \Rightarrow {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}} = {\log _{{3^{ - 1}}}}\left( 3 \right)\]
Using the logarithmic rule \[{\log _{\dfrac{1}{a}}}a = - {\log _a}a\], we get
\[ \Rightarrow {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}} = - {\log _3}\left( 3 \right)\]
By using the logarithmic rule \[{\log _a}a = - 1\], we get
\[ \Rightarrow {\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}} = - 1\]

Therefore, \[{\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}}\] is \[1\].

Note:
We know that a logarithmic equation is an equation that involves the logarithm of an expression with a variable on either of the sides. 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.