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

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

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

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

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}`