Question

How do I solve for x:

(4/9)^(x) • (8/27)^(1-x) =2/3

Answer 1

X=2

All this stuff is wrong I screwed up somewhere:
IGNORE EVERYTHING BELOW THIS:
take the natural log of both sides of the equation.

$ln(( \frac{4}{9})^{x} ) \times \ ln(( \frac{8}{27})^{(1 - x)} ) = ln( \frac{2}{3} )$
Remembering your log rules:

$ln( {a}^{b} ) = b \: ln(a)$
Using this rule we rewrite the first equation:
$x \: ln( \frac{4}{9} ) \times (1 - x) \: ln( \frac{8}{27} ) = ln( \frac{2}{3} )$
Simplify:
$x ln( \frac{4}{9} ) \times ( ln( \frac{8}{27}) - x \: ln( \frac{8}{27} ) )$
Solve for some of these ln's to make life easier:

$- .81x \: \times ( - 1.22 - 1.22x) = - .41$
Distribute:
$(.988x + .988x^{2} ) = - .41$

Factor out .988
$.988(x + {x}^{2} ) = - .41 \\ (x + {x}^{2} ) = \frac{ - .41}{.988} \\ (x + {x}^{2} ) = - .415$

Answer 2

$(\frac{4}{9})^x\cdot (\frac{8}{27})^{1-x} =\frac{2}{3}\\((\frac{2}{3})^2)^x\cdot ((\frac{2}{3})^3)^{1-x} =\frac{2}{3}\\(\frac{2}{3})^{2x}\cdot (\frac{2}{3})^{3-3x} =\frac{2}{3}\\(\frac{2}{3})^{2x+3-3x} =\frac{2}{3}\\(\frac{2}{3})^{-x+3} =\frac{2}{3}\\-x+3=1\\-x=1-3\\-x=-2\ /:(-1)\\x=2$