Question

Help with Algebra 2: 1. $\frac{\sqrt[7]{x^5} }{\sqrt[4]{x^2} }$
2.$3\sqrt[4]{(x-2)^3} -4=20$

Answer 1

Answer:

See explanation

Step-by-step explanation:

1. Given the expression

$\dfrac{\sqrt[7]{x^5} }{\sqrt[4]{x^2} }$

Note that

$\sqrt[7]{x^5}=x^{\frac{5}{7}} \\ \\\sqrt[4]{x^2}=x^{\frac{2}{4}}=x^{\frac{1}{2}}$

When dividing $\sqrt[7]{x^5}$ by $\sqrt[4]{x^2},$ we have to subtract powers (we cannot subtract 4 from 7, because then we get another expression), so

$\dfrac{5}{7}-\dfrac{2}{4}=\dfrac{5}{7}-\dfrac{1}{2}=\dfrac{5\cdot 2-1\cdot 7}{14}=\dfrac{3}{14}$

and the result is $x^{\frac{3}{14}}=\sqrt[14]{x^3}$

2. Given equation $3\sqrt[4]{(x-2)^3} -4=20$

Add 4:

$3\sqrt[4]{(x-2)^3} -4+4=20+4\\ \\3\sqrt[4]{(x-2)^3}=24$

Divide by 3:

$\sqrt[4]{(x-2)^3} =8$

Rewrite the equation as:

$(x-2)^{\frac{3}{4}}=8\\ \\(x-2)^{\frac{3}{4}}=2^3$

Hence,

$\left((x-2)^{\frac{3}{4}}\right)^{\frac{4}{3}}=(2^3)^{\frac{4}{3}}\\ \\x-2=2^{3\cdot \frac{4}{3}}\\ \\x-2=2^4\\ \\x-2=16\\ \\x-2+2=16+2\\ \\x=18$