Question 716905
{{{x^(-2) / (2(y^(-1))^2)}}}
Of course must follow the order of operations (aka PEMDAS or GEMDAS). So we start with the power of a power in the denominator. The rule here is to multiply the exponents:
{{{x^(-2) / (2y^(-2))}}}
This is a simplified expression... unless you consider negative exponents "unsimplified". If this is so then we can eliminate the negative exponents quickly or slowly.<br>
Quickly.
The fast way is to understand that negative exponents mean reciprocals. A negative exponent on a factor in the numerator becomes that factor with a positive exponent in the denominator and vice versa. Using this
{{{x^(-2) / (2y^(-2))}}}
becomes:
{{{y^2 / (2x^2)}}}
This is the simplified expression with positive exponents. (Note: Exponents only apply to whatever is immediately in front of it! So the exponent of -2 in the denominator applies only to the y, not to the 2! This is why the 2 stays in the denominator.)<br>
Slowly.
The methodical way is to use the rule:
{{{a^(-n) = 1/a^n}}}
Using this rule on the two negative exponents in:
{{{x^(-2) / (2y^(-2))}}}
we get:
{{{(1/x^2)/(2(1/y^2)))}}}
Simplifying...
{{{(1/x^2)/(2/y^2))}}}
Multiplying the numerator and denominator of the "big" fraction by the lowest common denominator of the "little" fractions, {{{x^2y^2}}}, will eliminate the "little" fractions:
{{{((1/x^2)/(2/y^2))*((x^2y^2)/(x^2y^2))}}}
giving us:
{{{y^2 / (2x^2)}}}