Question

Solve each equation (quadratic pattern)

$2^{2x} -2^{x} =12$

$3^{2x} + 3^{x+1} =4$

$4^{x} + 6$ · $2^{x} +8 = 0$

$9^{x} = 3^{x} +6$

Answer 1

Answer:  x = 2

Step-by-step explanation:

$2^{2x}-2^x-12=0\\\\\text{Let u = }2^x\\\\u^2-u-12=0\\(u-4)(u+3)=0\\\\u-4=0\quad and\quad u+3=0\\u=4\qquad and\quad u=-3\\\\\text{Substitute u with }2^x\\2^x=4\qquad and \quad 2^x=-3\\2^x=2^2\quad and\quad \text{not possible}\\\boxed{x=2}$

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Answer:  x = 0

Step-by-step explanation:

$3^{2x}+3^{x+1}-4=0\\\\3^{2x}+3^x\cdot3^1-4=0\\\\\text{Let u = }3^x\\u^2+3u-4=0\\\\(u+4)(u-1)=0\\\\u+4=0\quad and\quad u-1=0\\u=-4\qquad and\quad u=1\\\\\text{Substitute u with }3^x\\3^x=-4\qquad and\quad 3^x=1\\\text{not possible}\ and\quad 3^x=3^0\\.\qquad \qquad \qquad \qquad \boxed{x=0}$

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Answer:  No Solution

Step-by-step explanation:

$4^x+6\cdot 2^x+8=0\\\\2\cdot 2^x+6\cdot 2^x+8=0\\\\\text{Let u = }2^x\\2u+6u+8=0\\8u+8=0\\8u=-8\\u=-1\\\\\text{Substitute u with }2^x\\2^x=-1\\\text{not possible}$

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Answer:  No Solution

Step-by-step explanation:

$9^x=3^x-6\\\\3\cdot 3^x=1\cdot 3^x-6\\\\\text{Let u = }3^x\\\\3u=u-6\\2u=-6\\u=-3\\\\\text{Substitute u with }3^x\\3^x=-3\\\text{not possible}$