Question
Solve for $x$ :$(49)^{x + 4}= 7^2 \times(343)^{x + 1}$
✓
Answer
$(49)^{x + 4}= 7^2\times (343)^{x + 1}$
$\Rightarrow ( 7 \times 7 )^{x + 4} = 7^2 ( 7 \times 7 \times 7 )^{( x + 1 )}$
$\Rightarrow ( 7^2 )^{x + 4} = 7^2( 7^3 )^{( x + 1 )}$
$\Rightarrow 7^{( 2x + 8 )} = 7^2 \times 7^{3x + 3}$
$\Rightarrow 7^{( 2x + 8 )} = 7^{3x + 3 + 2}$
$\Rightarrow 7^{( 2x + 8 )} = 7^{3x + 5}$
We know that if bases are equal, the powers are equal
$\Rightarrow 2x + 8 = 3x + 5$
$\Rightarrow 3x - 2x = 8 - 5$
$\Rightarrow x = 3$
$\Rightarrow ( 7 \times 7 )^{x + 4} = 7^2 ( 7 \times 7 \times 7 )^{( x + 1 )}$
$\Rightarrow ( 7^2 )^{x + 4} = 7^2( 7^3 )^{( x + 1 )}$
$\Rightarrow 7^{( 2x + 8 )} = 7^2 \times 7^{3x + 3}$
$\Rightarrow 7^{( 2x + 8 )} = 7^{3x + 3 + 2}$
$\Rightarrow 7^{( 2x + 8 )} = 7^{3x + 5}$
We know that if bases are equal, the powers are equal
$\Rightarrow 2x + 8 = 3x + 5$
$\Rightarrow 3x - 2x = 8 - 5$
$\Rightarrow x = 3$
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