Question

Simplify the following: A. (-x)^4/4x * 8(-x)^-3/x^-3/4

Answer 1

Answer:

a. $-2x\sqrt[3]{x}$

b. $\frac{1}{2^x}$

Step-by-step explanation:

a.

Original equation:

$\frac{(-x)^4}{4x}*\frac{8(-x)^{-3}}{x^{-\frac{4}{3}}}$

So (-x)^4 can be seen as (-x * -x) * (-x * -x), which becomes x^2 * x^2 = x^4, the negatives cancel out of the degree is even. So it becomes:

$\frac{x^4}{4x}*\frac{8(-x)^{-3}}{x^{-\frac{4}{3}}}$

Cancel out one of the x's on the left fraction:

$\frac{x^3}{4}*\frac{8(-x)^{-3}}{x^{-\frac{4}{3}}}$

Rewrite the exponent in the numerator: $a^{-x} = \frac{1}{a^x}$

$\frac{x^3}{4}*\frac{8*\frac{1}{-x^3}}{x^{-\frac{4}{3}}}$

Simplify the numerator:

$\frac{x^3}{4}*\frac{\frac{8}{-x^3}}{x^{-\frac{4}{3}}}$

Keep numerator, change division to multiplication, flip the denominator:

$\frac{x^3}{4}*\frac{8}{-x^3} * \frac{1}{x^{-\frac{4}{3}}}$

multiply the denominator using the exponent identity: $x^a*x^b=x^{a+b}$

$\frac{x^3}{4}*\frac{8}{-x^{\frac{5}{3}}}$

Multiply the numerators and denominators:

$\frac{8x^3}{-4x^{\frac{5}{3}}}$

Use the fact that: $\frac{x^a}{x^b}=x^{a-b}$ to divide the x^3 and x^(5/3) and divide the 4 by the -8

$-2x^{\frac{4}{3}}$

Rewrite the exponent using the exponent identity: $x^{\frac{a}{b}} = \sqrt[b]{x^a}=\sqrt[b]{x}^a$

$-2\sqrt[3]{x^4}$

Rewrite as two radicals: $\sqrt[n]{a} * \sqrt[n]{b} = \sqrt[n]{ab}$

$-2\sqrt[3]{x^3} * \sqrt[3]{x}$

Simplify:

$-2x\sqrt[3]{x}$

b.

$2^{2x}\div4^{3x}*64^{\frac{x}{2}}$

Rewrite the 4 as 2^2

$2^{2x}\div(2^2)^{3x}*64^{\frac{x}{2}}$

Use the exponent identity: $(x^a)^b=x^{ab}$

$2^{2x}\div2^{6x}}*64^{\frac{x}{2}}$

Use the exponent identity: $\frac{x^a}{x^b}=x^{a-b}$

$2^{2x-6x} = 2^{-4x}$

Rewrite this part using the definition of a negative exponent: $(\frac{a}{b})^{-x} = \frac{b}{a^x}$.

$\frac{1}{2^{4x}} * 64^{\frac{x}{2}}$

Multiply:

$\frac{64^{\frac{x}{2}}}{2^{4x}}$

rewrite 64 as 2^6

$\frac{(2^6)^{\frac{x}{2}}}{2^{4x}}$

Use the identity: $(x^a)^b=x^{ab}$

$\frac{2^{3x}}{2^{4x}}$

Use the identity: $\frac{x^a}{x^b}=x^{a-b}$

$2^{-x}$

rewrite using the definition of a negative exponent: $(\frac{a}{b})^{-x} = \frac{b}{a^x}$

$\frac{1}{2^x}$

Answer 2

Answer:

Answer for (a)   -2x^4/3

Answer for (b)  2^5x/4^3x