QuestionAugust 26, 2025

5. 3^-2cdot 3^3x-7=((1)/(27))^2x+15

5. 3^-2cdot 3^3x-7=((1)/(27))^2x+15
5. 3^-2cdot 3^3x-7=((1)/(27))^2x+15

Solution
3.4(258 votes)

Answer

x = -4 Explanation 1. Simplify the left side Combine exponents: 3^{-2} \cdot 3^{3x-7} = 3^{-2 + 3x - 7} = 3^{3x - 9}. 2. Simplify the right side Convert to base 3: \left(\frac{1}{27}\right)^{2x+15} = (3^{-3})^{2x+15} = 3^{-3(2x+15)} = 3^{-6x - 45}. 3. Set exponents equal Since bases are equal, set exponents equal: 3x - 9 = -6x - 45. 4. Solve for x Rearrange and solve: \[ 3x + 6x = -45 + 9 \] \[ 9x = -36 \] \[ x = -4 \]

Explanation

1. Simplify the left side<br /> Combine exponents: $3^{-2} \cdot 3^{3x-7} = 3^{-2 + 3x - 7} = 3^{3x - 9}$.<br /><br />2. Simplify the right side<br /> Convert to base 3: $\left(\frac{1}{27}\right)^{2x+15} = (3^{-3})^{2x+15} = 3^{-3(2x+15)} = 3^{-6x - 45}$.<br /><br />3. Set exponents equal<br /> Since bases are equal, set exponents equal: $3x - 9 = -6x - 45$.<br /><br />4. Solve for x<br /> Rearrange and solve: <br />\[ <br />3x + 6x = -45 + 9 <br />\]<br />\[ <br />9x = -36 <br />\]<br />\[ <br />x = -4 <br />\]
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