Question

6 solve for x:
★ ( x^{-1} = \frac{1}{6} )
( \boldsymbol{\text{b}} ) ( x^{-2} = \frac{1}{9} )
( \boldsymbol{\text{c}} ) ( x^{-3} = -\frac{1}{27} )
( \boldsymbol{\text{d}} ) ( x^{-2} = 49 )
( \boldsymbol{\text{e}} ) ( x^{-4} = \frac{1}{16} )
★ ( x^{-3} = -64 )
( \boldsymbol{\text{g}} ) ( (x + 1)^{-2} = -4 )
★ ( (2x - 5)^{-3} = \frac{1}{5} )

Explanation:

Step1: Rewrite negative exponents

Recall $x^{-n}=\frac{1}{x^n}$ for each equation.

Step2: Solve for $x$ (case by case)

For a: $x^{-1}=\frac{1}{6}$

Rewrite: $\frac{1}{x}=\frac{1}{6}$
Cross-multiply: $x=6$

For b: $x^{-2}=\frac{1}{9}$

Rewrite: $\frac{1}{x^2}=\frac{1}{9}$
Take reciprocal: $x^2=9$
Take square root: $x=\pm3$

For c: $x^{-3}=-\frac{1}{27}$

Rewrite: $\frac{1}{x^3}=-\frac{1}{27}$
Take reciprocal: $x^3=-27$
Take cube root: $x=-3$

For d: $x^{-2}=49$

Rewrite: $\frac{1}{x^2}=49$
Take reciprocal: $x^2=\frac{1}{49}$
Take square root: $x=\pm\frac{1}{7}$

For e: $x^{-4}=\frac{1}{16}$

Rewrite: $\frac{1}{x^4}=\frac{1}{16}$
Take reciprocal: $x^4=16$
Take 4th root: $x=\pm2$

For f: $x^{-3}=-64$

Rewrite: $\frac{1}{x^3}=-64$
Take reciprocal: $x^3=-\frac{1}{64}$
Take cube root: $x=-\frac{1}{4}$

For g: $(x+1)^{-2}=-4$

Rewrite: $\frac{1}{(x+1)^2}=-4$
A square of a real number is non-negative, so $\frac{1}{(x+1)^2}>0$. No real solution exists.

For h: $(2x-5)^{-3}=\frac{1}{5}$

Rewrite: $\frac{1}{(2x-5)^3}=\frac{1}{5}$
Take reciprocal: $(2x-5)^3=5$
Take cube root: $2x-5=\sqrt[3]{5}$
Solve for $x$: $x=\frac{5+\sqrt[3]{5}}{2}$

Answer:

a. $x=6$
b. $x=3$ or $x=-3$
c. $x=-3$
d. $x=\frac{1}{7}$ or $x=-\frac{1}{7}$
e. $x=2$ or $x=-2$
f. $x=-\frac{1}{4}$
g. No real solution
h. $x=\frac{5+\sqrt[3]{5}}{2}$