Answer to Question #223002 in Algebra for Cheryl-Ann

Question

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

Answer 1

Let us solve the equation

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

which is equivalent to the following equations:

$\frac{2^{2(3+x)}}{2^{30x}} = \frac{2^{10-2x}}{2^{18x}}$

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

$2^{6-28x} = 2^{10-20x}$

$6-28x= 10-20x$

$8x= -4$

$x=-\frac{1}{2}$

We conclude that the value of $x$ is $-\frac{1}{2}.$