3 years ago

Solve for the value of X in 9^x+4(3^x+1)+27=0

1 answer:

3 0

Answer:

No solutions

Step-by-step explanation:

$9^x + 4(3^x + 1) + 27 = 0\\3^{2x} + 4 \cdot 3^x + 31 = 0$

Substitute 3^x with t and we get.

$t^2 + 4t + 31 = 0\\\triangle = 16 - 4 \cdot 31 = -108 < 0 => t_{12} \notin R => x \notin R$

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Solve for the value of X in 9^x+4(3^x+1)+27=0​