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I am trying to find the value of x in the following problem, I have to solve it without logarithm.

Problem : $ \dfrac {27 ^ {(2x+1)} } { 3 ^ {(x+1){5}}} = \dfrac{1}{3} $

EDIT: My work so far:

$ \dfrac {3^{3(2x+1)} } { 3 ^ {(x+1)5}} = 3^{-1} $

I know the formula $ b^{u} = b^{v} <=> u = v $ but I am not able to use it with this problem.

Thanks for help !

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    Additionally, $\frac{3^a}{3^b}=3^{a-b}$ and $\frac1{3^a}=3^{-a}$...2011-11-11

1 Answers 1

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Let's combine all the hints (from above comments) to write:

$\dfrac{27^{(2x + 1)}}{3^{5(x + 1)}} = \dfrac{(3^3)^{(2x + 1)}}{3^{(5x + 5)}} = \dfrac{3^{3(2x + 1)}}{3^{(5x + 5)}} = \dfrac{3^{(6x + 3)}}{3^{(5x + 5)}} = 3^{(6x + 3) -(5x + 5)} = 3^{-1}$

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    There you go! And you can always check your solution by plugging $x = 1$ into the original equation. And welcome to math.stackexchange!2011-11-11