Find value of x: $$\frac{27^x}{9^{2x-1}}=3^{x+4}$$
My steps: $$\frac{(3^3)^x}{(3^2)^{2x-1}}=3^{x+4}$$ $$\frac{3x}{4x-2}=x+4$$
Please help me finish solving, and correct me if what I did so far has mistakes. Thanks very much.
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2$$\frac{a^m}{a^n}=a^{m-n} \ \neq \frac{m}{n}$$ your mistake is this $$\frac{3^{3x}}{3^4x-2}=3^{(3x)-(4x-2)}\\neq \frac{3x}{4x-2} $$ – Khosrotash Jul 09 '15 at 18:38
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@daryakhosrotash I think you meant $\ne a^{\frac mn}$. – peterwhy Jul 09 '15 at 18:41
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no,exactly I try to show "Arthur Alex Karapetov" mistake – Khosrotash Jul 09 '15 at 18:44
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@daryakhosrotash I don't think OP claimed $$\dfrac {a^m}{a^n} = \dfrac mn$$ just as I suppose OP does not think that $3^{x+4} = x+4$. What happened from line 2 to 3 was a wrong simplification using $$\dfrac {a^m}{a^n} = a^\frac mn$$ and then take $\log_3$ simultaneously. – peterwhy Jul 09 '15 at 19:00
