How do you solve ${(\sqrt{3})}^{2 x + 4} = 9^{x - 2}$ $(sqrt(3))^(2x+4)=9^(x-2)$ ?

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Answer 1 · 2015-10-09T16:30:04

X = 6

${(\sqrt{3})}^{2 X + 4}$ $(sqrt3)^(2X+4)$ = $9^{X - 2}$ $9^(X-2)$

$3^{\frac{1}{2} (2 X + 4)}$ $3^(1/2(2X+4)$ = $3^{2 (X - 2)}$ $3^(2(X-2)$
$3^{(X + 2)}$ $3^((X+2)$ = $3^{2 X - 4}$ $3^(2X-4)$

Take ln both sides

$ln (3^{(X + 2)})$ $ln(3^((X+2)))$ = $ln (3^{2 X - 4})$ $ln(3^(2X-4))$

X+2 (ln3) = 2X-4 ( ln3)

ln3 on both side cancels,
X+2 = 2X-4
Rearrange terms

2+4 = 2X-X

6 =- X

How do you solve (sqrt(3))^(2x+4)=9^(x-2)(√3)2x+4=9x−2 (sqrt(3))^(2x+4)=9^(x-2)?