Verified answer

Answer:

[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \:(1) \: \: log\dfrac{33}{45} + log\dfrac{490}{297} - 2log\dfrac{21}{27} = log2 \qquad \: \\ \\& \qquad \:\sf \:(2) \: \: log_{10}(0.001) = \: - \: 3 \\ \\& \qquad \:\sf \:(3) \: \: xyz \: = \: \pm \: 12 \end{aligned}} \qquad \\ \\ [/tex]

Step-by-step explanation:

[tex]\large\underline{\sf{Solution-1}}[/tex]

Given expression is

[tex]\sf \: log\dfrac{33}{45} + log\dfrac{490}{297} - 2log\dfrac{21}{27} \\ \\ [/tex]

[tex]\sf \: log\dfrac{11}{15} + log\dfrac{490}{297} - 2log\dfrac{7}{9} \\ \\ [/tex]

[tex]\sf \: = \: log\dfrac{11}{15} + log\dfrac{490}{297} - log\left(\dfrac{7}{9}\right) ^{2} \\ \\ [/tex]

[tex]\sf \: = \: log\dfrac{11}{15} + log\dfrac{490}{297} - log\dfrac{49}{81} \\ \\ [/tex]

[tex]\sf \: = \: log\left(\dfrac{11}{15} \times \dfrac{490}{297} \times \dfrac{81}{49}\right) \\ \\ [/tex]

[tex]\sf \: = \: log\left(\dfrac{1}{15} \times \dfrac{490}{27} \times \dfrac{81}{49}\right) \\ \\ [/tex]

[tex]\sf \: = \: log\left(\dfrac{1}{15} \times \dfrac{490}{1} \times \dfrac{3}{49}\right) \\ \\ [/tex]

[tex]\sf \: = \: log\left(\dfrac{1}{15} \times 10 \times \dfrac{3}{1}\right) \\ \\ [/tex]

[tex]\sf \: = \: log2 \\ \\ [/tex]

Hence,

[tex]\sf\implies \bf \: log\dfrac{33}{45} + log\dfrac{490}{297} - 2log\dfrac{21}{27} = log2 \\ \\ \\ [/tex]

[tex]\large\underline{\sf{Solution-2}}[/tex]

Given expression is

[tex]\sf \: log_{10}(0.001) \\ \\ [/tex]

[tex]\sf \: = \: log_{10}\left(\dfrac{1}{1000} \right) \\ \\ [/tex]

[tex]\sf \: = \: log_{10}\left(\dfrac{1}{ {10}^{3} } \right) \\ \\ [/tex]

[tex]\sf \: = \: log_{10}\left( {10}^{ - 3} \right) \\ \\ [/tex]

[tex]\sf \: = \: - \: 3 \\ \\ [/tex]

Hence,

[tex]\sf\implies \bf \: log_{10}(0.001) = \: - \: 3 \\ \\ \\ [/tex]

[tex]\large\underline{\sf{Solution-3}}[/tex]

Given that,

[tex]\sf \: {x}^{2} =3 \\ \\ [/tex]

[tex]\sf \: {y}^{2} = 6 \\ \\ [/tex]

[tex]\sf \: {z}^{2} = 8 \\ \\ [/tex]

On multiply above 3 equations, we get

[tex]\sf \: {x}^{2} \times {y}^{2} \times {z}^{2} = 3 \times 6 \times 8 \\ \\ [/tex]

[tex]\sf \: {x}^{2}{y}^{2} {z}^{2} = 3 \times 3 \times 2 \times 2 \times 2 \times 2 \\ \\ [/tex]

[tex]\sf \: {(xyz)}^{2} = 3 \times 3 \times 2 \times 2 \times 2 \times 2 \\ \\ [/tex]

[tex]\sf \: xyz \: = \: \pm \: \sqrt{3 \times 3 \times 2 \times 2 \times 2 \times 2} \\ \\ [/tex]

[tex]\sf \: xyz \: = \: \pm \: 3 \times 2 \times 2 \\ \\ [/tex]

[tex]\sf\implies \bf \: xyz \: = \: \pm \: 12 \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Formulae Used

[tex]\sf \: log(xy) = logx + logy \\ \\ [/tex]

[tex]\sf \: log( {x}^{y}) = ylogx \\ \\ [/tex]

[tex]\sf \: log\bigg(\dfrac{x}{y} \bigg) = logx - logy \\ \\ [/tex]

[tex]\sf \: log_{a}( {a}^{b} ) = b \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ log_{x}(x) = 1}\\ \\ \bigstar \: \bf{ log_{x}( {x}^{y} ) = y}\\ \\ \bigstar \: \bf{ log_{ {x}^{z} }( {x}^{w} ) = \dfrac{w}{z} }\\ \\ \bigstar \: \bf{ log_{a}(b) = \dfrac{logb}{loga} }\\ \\ \bigstar \: \bf{ {e}^{logx} = x}\\ \\ \bigstar \: \bf{ {e}^{ylogx} = {x}^{y}}\\ \\ \bigstar \: \bf{log1 = 0}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]