Question

2. Simplify 92(1095 3) + 27log6 36 + 6/3log79​

Answer 1

Answer:

$\: \boxed{\bf \: 9^{\frac{1}{2 log_{5}(3) }} + {27}^{ log_{6}(36) } + {3}^{ \frac{6}{ log_{7}(9) } } \: = \: 1077 \: }$

Step-by-step explanation:

Given logarithmic expression is

$\sf \: 9^{\frac{1}{2 log_{5}(3) }} + {27}^{ log_{6}(36) } + {3}^{ \frac{6}{ log_{7}(9) } }$

can be rewritten as

$\sf \: = \: 3^{2 \times \frac{1}{2 log_{5}(3) }} - {27}^{ log_{6}( {6}^{2} ) } + {3}^{ \frac{6}{ log_{7}(9) } }$

We know,

$\boxed{\begin{aligned}& \qquad \:\sf \: log_{x}(y) = \dfrac{1}{ log_{y}(x) } \qquad \: \\ \\& \qquad \:\sf \: log_{ {x}^{m} }( {y}^{n} ) = \frac{n}{m} log_{x}(y) \end{aligned}} \qquad$

So, using this property, the above expression can be rewritten as

$\sf \: = \: 3^{\frac{1}{log_{5}(3) }} + {27}^{ 2} + {3}^{6 log_{9}(7) }$

can be rewritten as

$\sf \: = \: 3^{ log_{3}(5) } + 729 + {3}^{6 log_{ {3}^{2} }(7) }$

We know,

$\boxed{\sf \: {a}^{ log_{a}(x) } = x \: }$

So, using this result, the above expression can be rewritten as

$\sf \: = \: 5 + 729 + {3}^{6 \times \frac{1}{2} log_{3}(7) }$

$\sf \: = \: 734 + {3}^{3 log_{3}(7) }$

$\sf \: = \: 734 + {3}^{ log_{3}( {7}^{3} ) }$

$\sf \: = \: 734 + {7}^{3}$

$\sf \: = \: 734 + 343$

$\sf \: = \: 1077$

Hence,

$\implies\sf \: \boxed{\bf \: 9^{\frac{1}{2 log_{5}(3) }} + {27}^{ log_{6}(36) } + {3}^{ \frac{6}{ log_{7}(9) } } \: = \: 1077 \: }$

$\rule{190pt}{2pt}$

Additional Information

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