QuestionJune 5, 2025

((1)/(27))^x^(2)=(9^3x+8)((1)/(81))^x^(2)

((1)/(27))^x^(2)=(9^3x+8)((1)/(81))^x^(2)
((1)/(27))^x^(2)=(9^3x+8)((1)/(81))^x^(2)

Solution
4.1(230 votes)

Answer

No real solutions exist. Explanation 1. Simplify the bases Rewrite 9 as 3^2, 27 as 3^3, and 81 as 3^4. The equation becomes (3^{-3})^{x^2} = (3^{6x+16})(3^{-4x^2}). 2. Apply exponent rules Use **(a^m)^n = a^{mn}** to simplify: 3^{-3x^2} = 3^{6x + 16 - 4x^2}. 3. Equate exponents Since bases are equal, equate exponents: -3x^2 = 6x + 16 - 4x^2. 4. Solve for x Rearrange: x^2 + 6x + 16 = 0. Factor or use quadratic formula **x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}**. Here, a=1, b=6, c=16. Calculate discriminant: b^2 - 4ac = 36 - 64 = -28. No real solutions.

Explanation

1. Simplify the bases<br /> Rewrite $9$ as $3^2$, $27$ as $3^3$, and $81$ as $3^4$. The equation becomes $(3^{-3})^{x^2} = (3^{6x+16})(3^{-4x^2})$.<br />2. Apply exponent rules<br /> Use **$(a^m)^n = a^{mn}$** to simplify: $3^{-3x^2} = 3^{6x + 16 - 4x^2}$.<br />3. Equate exponents<br /> Since bases are equal, equate exponents: $-3x^2 = 6x + 16 - 4x^2$.<br />4. Solve for $x$<br /> Rearrange: $x^2 + 6x + 16 = 0$. Factor or use quadratic formula **$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$**. Here, $a=1$, $b=6$, $c=16$. Calculate discriminant: $b^2 - 4ac = 36 - 64 = -28$. No real solutions.
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