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Formula
Solve the equation
Graph
$y = 3 ^ { 2 } \times 81$
$y = 3 ^ { x + 1 }$
$y$Intercept
$\left ( 0 , 3 \right )$
Asymptote
$y = 0$
$3 ^{ 2 } \times 81 = 3 ^{ x+1 }$
$x = 5$
Solve the equation
$\color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ \times } \color{#FF6800}{ 81 } = \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } }$
 Invert the left and right terms to solve the exponential equation (inequality) 
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \log _{ \color{#FF6800}{ 3 } } { \left( \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ \times } \color{#FF6800}{ 81 } \right) }$
$x + 1 = \log _{ 3 } { \left( \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ \times } \color{#FF6800}{ 81 } \right) }$
 Simplify the expression 
$x + 1 = \log _{ 3 } { \left( \color{#FF6800}{ 9 } \color{#FF6800}{ \times } \color{#FF6800}{ 81 } \right) }$
$x + 1 = \log _{ 3 } { \left( \color{#FF6800}{ 9 } \color{#FF6800}{ \times } \color{#FF6800}{ 81 } \right) }$
 Multiply $9$ and $81$
$x + 1 = \log _{ 3 } { \left( \color{#FF6800}{ 729 } \right) }$
$x + 1 = \log _{ 3 } { \left( \color{#FF6800}{ 729 } \right) }$
 Write the number in exponential form with base $3$
$x + 1 = \log _{ 3 } { \left( \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 6 } } \right) }$
$x + 1 = \log _{ \color{#FF6800}{ 3 } } { \left( \color{#FF6800}{ 3 } ^ { \color{#FF6800}{ 6 } } \right) }$
 Simplify the expression using $\log_{a}{a^{x}}=x\times\log_{a}{a}$
$x + 1 = \color{#FF6800}{ 6 } \log _{ \color{#FF6800}{ 3 } } { \left( \color{#FF6800}{ 3 } \right) }$
$x + 1 = 6 \log _{ \color{#FF6800}{ 3 } } { \left( \color{#FF6800}{ 3 } \right) }$
 The logarithm is equal to 1 if a base is same as an antilogarithm 
$x + 1 = 6 \times \color{#FF6800}{ 1 }$
$x + 1 = 6 \color{#FF6800}{ \times } \color{#FF6800}{ 1 }$
 Multiplying any number by 1 does not change the value 
$x + 1 = \color{#FF6800}{ 6 }$
$x \color{#FF6800}{ + } \color{#FF6800}{ 1 } = 6$
 Move the constant to the right side and change the sign 
$x = 6 \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$x = \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
 Subtract $1$ from $6$
$x = \color{#FF6800}{ 5 }$
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