Answer to Question #223297 in Algebra for George

Question

Given that 4^3+x/8^10x = 2^10-2x/64^3x what is the value of x?

Answer 1

$\frac{4^{3+x}}{8^{10x}} = \frac{2^{10-2x}} {64^{3x}}$

=> $4^{3+x} • 64^{3x} = 8^{10x} • 2^{10-2x}$

$2^{6+2x} • 2^{18x} = 2^{30x} • 2^{10-2x}$

$2^{6+2x+18x} = 2^{30x+10-2x}$

=> $6 + 2x +18x = 30x + 10 - 2x$

$20x - 28x = 10 - 6$

$-8x = 4$

=> $x = \frac{4}{-8} = \frac{-1}{2}$