Simplify: \[{\log _{\dfrac{1}{3}}}\sqrt[4]{{729\sqrt[3]{{{9^{ - 1}}{{27}^{\dfrac{{ - 4}}{3}}}}}}}\]
Answer
629.4k+ views
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.
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.
Recently Updated Pages
Find the greatest six digit number that is exactly class 8 maths CBSE

What is the time difference between India and Cana class 8 social science CBSE

Compare LPG and wood as fuels class 8 chemistry CBSE

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

30 opposite words in English from a to z class 8 english CBSE

How many cubic feet equals to 1 unit sand class 8 maths CBSE

Trending doubts
What is BLO What is the full form of BLO class 8 social science CBSE

Citizens of India can vote at the age of A 18 years class 8 social science CBSE

Full form of STD, ISD and PCO

One cusec is equal to how many liters class 8 maths CBSE

Who commanded the Hector the first British trading class 8 social science CBSE

What are the methods of reducing friction. Explain


