Question

The value of ${\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729 \cdot \sqrt[3]{{{9^{ - 1}} \cdot {{27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right)$
1.$-1$
2.1
3.2
4.Zero

Answer 1

Here, we will first convert all the exponents in fractional form. Then we will use the basic properties of exponents for all the exponentials inside the bracket and here we will also use the property of logarithms to simplify the given expressions further.

Complete step-by-step answer:
Here, we have to find the value of the given expression ${\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729 \cdot \sqrt[3]{{{9^{ - 1}} \cdot {{27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right)$
We can write 729 as ${3^6}$ and 27 as ${3^3}$and 9 as ${3^2}$.
${\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729 \cdot \sqrt[3]{{{9^{ - 1}} \cdot {{27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right) = {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {{3^6} \cdot \sqrt[3]{{{{\left( {{3^2}} \right)}^{ - 1}} \cdot {{\left( {{3^3}} \right)}^{ - \dfrac{4}{3}}}}}} }}{{{{\left( {{2^2}} \right)}^{{{\log }_2}3}}}}} \right)$
Now, we will multiply the powers of exponentials. Therefore, we get
$\begin{array}{l} \Rightarrow {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729 \cdot \sqrt[3]{{{9^{ - 1}} \cdot {{27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right) = {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {{3^6} \cdot \sqrt[3]{{{3^2}^{ \times - 1} \cdot {3^{\left( {3 \times - \dfrac{4}{3}} \right)}}}}} }}{{{2^2}^{ \times {{\log }_2}3}}}} \right)\\\ \Rightarrow {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729 \cdot \sqrt[3]{{{9^{ - 1}} \cdot {{27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right) = {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {{3^6}.\sqrt[3]{{{3^{ - 2}}{{.3}^{ - 4}}}}} }}{{{2^2}^{ \times {{\log }_2}3}}}} \right)\end{array}$
We know when the powers with the same base are multiplied, their exponents get added $\begin{array}{l} \Rightarrow {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729 \cdot \sqrt[3]{{{9^{ - 1}} \cdot {{27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right) = {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {{3^6} \cdot \sqrt[3]{{{3^{ - 2 + \left( { - 4} \right)}}}}} }}{{{2^2}^{ \times {{\log }_2}3}}}} \right)\\\ \Rightarrow {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729 \cdot \sqrt[3]{{{9^{ - 1}} \cdot {{27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right) = {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {{3^6} \cdot \sqrt[3]{{{3^{ - 6}}}}} }}{{{2^2}^{ \times {{\log }_2}3}}}} \right)\end{array}$
Now, we will find the cube root of the exponential ${3^{ - 6}}$.
$\Rightarrow {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729 \cdot \sqrt[3]{{{9^{ - 1}} \cdot {{27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right) = {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {{3^6} \cdot {3^{ - 2}}} }}{{{2^2}^{ \times {{\log }_2}3}}}} \right)$
We know when the powers with the same base are multiplied, their exponents get added.
$\Rightarrow {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729 \cdot \sqrt[3]{{{9^{ - 1}} \cdot {{27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right) = {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {{3^6}^{ + \left( { - 2} \right)}} }}{{{2^2}^{ \times {{\log }_2}3}}}} \right)$
On further simplification, we get
$\begin{array}{l} \Rightarrow {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729 \cdot \sqrt[3]{{{9^{ - 1}} \cdot {{27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right) = {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {{3^6}^{ - 2}} }}{{{2^2}^{ \times {{\log }_2}3}}}} \right)\\\ \Rightarrow {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729 \cdot \sqrt[3]{{{9^{ - 1}} \cdot {{27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right) = {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {{3^4}} }}{{{2^2}^{ \times {{\log }_2}3}}}} \right)\end{array}$
Now, we will find the square root of the exponential ${3^4}$.
$\Rightarrow {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729 \cdot \sqrt[3]{{{9^{ - 1}} \cdot {{27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right) = {\log _{\dfrac{1}{3}}}\left( {\dfrac{{{3^2}}}{{{2^2}^{ \times {{\log }_2}3}}}} \right)$
We know one property of logarithm that ${\log _a}{b^n} = n{\log _a}b$ .
Therefore, using this property, we get
$\Rightarrow {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729 \cdot \sqrt[3]{{{9^{ - 1}} \cdot {{27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right) = {\log _{\dfrac{1}{3}}}\left( {\dfrac{{{3^2}}}{{{2^{{{\log }_2}{3^2}}}}}} \right)$
Now, we will again simplify the term in numerator using the property of logarithm.
We know the property of logarithm that ${\log _b}{b^n} = n$.
Therefore, using this property, we get
$\Rightarrow {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729 \cdot \sqrt[3]{{{9^{ - 1}} \cdot {{27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right) = {\log _{\dfrac{1}{3}}}\left( {\dfrac{{{3^2}}}{{{3^2}}}} \right)$
On further simplification, we get
$\Rightarrow {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729 \cdot \sqrt[3]{{{9^{ - 1}} \cdot {{27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right) = {\log _{\dfrac{1}{3}}}1$
We know the property of logarithm that ${\log _b}1 = 0$.
Therefore, we get
$\Rightarrow {\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729 \cdot \sqrt[3]{{{9^{ - 1}} \cdot {{27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right) = {\log _{\dfrac{1}{3}}}1 = 0$
Thus, the value of ${\log _{\dfrac{1}{3}}}\left( {\dfrac{{\sqrt {729.\sqrt[3]{{{9^{ - 1}}{{.27}^{ - \dfrac{4}{3}}}}}} }}{{{4^{{{\log }_2}3}}}}} \right)$ is equal to zero.
Hence, the correct option is option 4.

Note: Here, we need to know all the properties of exponentials and the basic properties of logarithms. The logarithm is defined as the inverse function of the exponential function. We need to keep in mind that when we take the inverse of the exponential function, we get the logarithmic function.
In addition to this, the power property of exponentials states that to find a power of a power we just need to multiply the exponents.