$3^{(2x+3)} - 2.9^{(x+1)} =1/3$
Please help me with this problem
Its my elementary mathematics indices problem
2026-03-26 20:36:41.1774557401
I got a problem in indices
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Using the index laws, transform $3^{2x+3} = 3^{2x} \times 3^3$ and $9^{x+1} = 3^{2(x+1)} = 3^{2x}\times 3^2$, so that the LHS becomes $3^{2x}(3^3 - 2\times 3^2) = 3^{2x}(3\times 3^2 - 2\times 3^2) = 3^{2x} \times 3^2$. Now you need only to solve $3^{2x} \times 3^2 = 3^{-1}$ which means $2x + 2 = -1$, so $x = -3/2$.