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$1 + 2 \log _{ 2 } { \left( 3 \right) }$
Calculate the value
$\log _{ 2 } { \left( \color{#FF6800}{ 18 } \right) }$
$ $ Factor the antilogarithm with the expression in which $ 2 $ , that is the base, is included $ $
$\log _{ 2 } { \left( \color{#FF6800}{ 2 } \right) } + \log _{ 2 } { \left( \color{#FF6800}{ 9 } \right) }$
$\log _{ \color{#FF6800}{ 2 } } { \left( \color{#FF6800}{ 2 } \right) } + \log _{ 2 } { \left( 9 \right) }$
$ $ The logarithm is equal to 1 if a base is same as an antilogarithm $ $
$\color{#FF6800}{ 1 } + \log _{ 2 } { \left( 9 \right) }$
$1 + \log _{ 2 } { \left( \color{#FF6800}{ 9 } \right) }$
$ $ Write the number in exponential form with base $ 3$
$1 + \log _{ 2 } { \left( \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } \right) }$
$1 + \log _{ \color{#FF6800}{ 2 } } { \left( \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } \right) }$
$ $ Simplify the expression using $ \log_{a}{b^{x}}=x\times\log_{a}{b}$
$1 + \color{#FF6800}{ 2 } \log _{ \color{#FF6800}{ 2 } } { \left( \color{#FF6800}{ 3 } \right) }$
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