# Find x in the following equation

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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Hint: $27=3^3$. – David Mitra Nov 11 '11 at 18:43
$$\frac{3^{3(2x+1)}}{3^{5(x+1)}}=3^{-1}$$ – pedja Nov 11 '11 at 18:45
Additionally, $\frac{3^a}{3^b}=3^{a-b}$ and $\frac1{3^a}=3^{-a}$... – J. M. Nov 11 '11 at 18:45

$\dfrac{27^{(2x + 1)}}{3^{5(x + 1)}} = \dfrac{(3^3)^{(2x + 1)}}{3^{(5x + 5)}} = \dfrac{3^{3(2x + 1)}}{3^{(5x + 5)}} = \dfrac{3^{(6x + 3)}}{3^{(5x + 5)}} = 3^{(6x + 3) -(5x + 5)} = 3^{-1}$
Is it $$6x+3-5x-5 = -1, x = 1$$ ? – Jonathan Nov 11 '11 at 18:57
Come on, now! You have to distribute that negative/minus!! So it would be $6x + 3 - 5x -5 = -1$ ... – The Chaz 2.0 Nov 11 '11 at 18:58
There you go! And you can always check your solution by plugging $x = 1$ into the original equation. And welcome to math.stackexchange! – The Chaz 2.0 Nov 11 '11 at 19:00