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$3^{(2x+3)} - 2.9^{(x+1)} =1/3$

Please help me with this problem

Its my elementary mathematics indices problem

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    $\begingroup$ Do you mean $2.9$ or $2\times9$? $\endgroup$ Jul 16, 2020 at 10:13
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    $\begingroup$ I am tempted to say it is $2\times9^{(x+1)}$, otherwise the solutions would be way more difficult, as logarithms would be involved. $\endgroup$
    – Logos
    Jul 16, 2020 at 10:47

1 Answer 1

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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 $x = -3/2$.

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