$\color{#FF6800}{ 9 } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 9 } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 9 } ^ { \color{#FF6800}{ 3 } } = \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ b } }$
$ $ Invert the left and right terms to solve the exponential equation (inequality) $ $
$\color{#FF6800}{ b } = \log _{ \color{#FF6800}{ 3 } } { \left( \color{#FF6800}{ 9 } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 9 } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 9 } ^ { \color{#FF6800}{ 3 } } \right) }$
$b = \log _{ 3 } { \left( \color{#FF6800}{ 9 } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 9 } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 9 } ^ { \color{#FF6800}{ 3 } } \right) }$
$ $ Add the forms of the powers with the same bases and exponents $ $
$b = \log _{ 3 } { \left( \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 9 } ^ { \color{#FF6800}{ 3 } } \right) }$
$b = \log _{ 3 } { \left( \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 9 } ^ { \color{#FF6800}{ 3 } } \right) }$
$ $ Simplify the expression $ $
$b = \log _{ 3 } { \left( \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 729 } \right) }$
$b = \log _{ 3 } { \left( \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 729 } \right) }$
$ $ Multiply $ 3 $ and $ 729$
$b = \log _{ 3 } { \left( \color{#FF6800}{ 2187 } \right) }$
$b = \log _{ 3 } { \left( \color{#FF6800}{ 2187 } \right) }$
$ $ Write the number in exponential form with base $ 3$
$b = \log _{ 3 } { \left( \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 7 } } \right) }$
$b = \log _{ \color{#FF6800}{ 3 } } { \left( \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 7 } } \right) }$
$ $ Simplify the expression using $ \log_{a}{a^{x}}=x\times\log_{a}{a}$
$b = \color{#FF6800}{ 7 } \log _{ \color{#FF6800}{ 3 } } { \left( \color{#FF6800}{ 3 } \right) }$
$b = 7 \log _{ \color{#FF6800}{ 3 } } { \left( \color{#FF6800}{ 3 } \right) }$
$ $ The logarithm is equal to 1 if a base is same as an antilogarithm $ $
$b = 7 \times \color{#FF6800}{ 1 }$
$b = 7 \color{#FF6800}{ \times } \color{#FF6800}{ 1 }$
$ $ Multiplying any number by 1 does not change the value $ $
$b = \color{#FF6800}{ 7 }$