I'm hoping someone can help me understand where I'm going wrong with this question.
Solve the equation: $$\frac{3^{5x+2}} {9^{1-x}} = \frac{27^{4+3x}}{729}$$
My working:
Make all terms powers of 3 $$\frac{3^{5x+2}} {(3^2)^{1-x}} = \frac{(3^3)^{4+3x}}{3^6}$$ Multiply out the indices $$\frac{3^{5x+2}} {3^{2-2x}} = \frac{3^{12+9x}}{3^6}$$ Work the indices only $$\frac{5x+2} {2-2x} = \frac{12+9x}{6}$$ Simplify as $-2.5x = 2+1.5x$ then $-5x =4 + 3x$ and so $-8x = 4$ and then $x = -1/2$
The book tells me the correct answer is $x = -3$
Thanks in advance
Gary
From $$\frac{3^{5x+2}} {3^{2-2x}} = \frac{3^{12+9x}}{3^6}$$ you work out the indices to get $$ 3^{(5x+2) - (2-2x) - (12+9x) + 6}=1=3^0, $$ and thus $$(5x+2) - (2-2x) - (12+9x) + 6=0$$
Can you finish this?