Question

Sort the given expressions in decreasing order.
(Highest on top)
Given:
$a=\frac{1}{3},b=-\frac{1}{3}$

• A. ${125}^b$
• B. ${64}^a$
• C. $(125+64{)}^{a+b}$
• D. ${64}^b+{125}^a$

Answer 1

Given:
$a=\frac{1}{3},b=-\frac{1}{3}$

${125}^b$ = ${125}^{-\frac{1}{3}}$ = $5^{3(-\frac{1}{3})}$ = $\frac{1}{5}$ = $0.2$


${64}^a$ = ${64}^{\frac{1}{3}}$ = $4^{3\left(\frac{1}{3}\right)}$ = 4


$(125+64{)}^{a+b}$ = $(189{)}^{a+b}$=$(189{)}^{\frac{1}{3}-\frac{1}{3}}$ = $(189{)}^0$ = 1


${64}^b+{125}^a$ = ${64}^{-\frac{1}{3}}$ + ${125}^{\frac{1}{3}}$ = $4^{3(-\frac{1}{3})}$ + $5^{3\left(\frac{1}{3}\right)}$ = $\frac{1}{4}$ + 5 = $5.25$

Hence,
(${64}^b+{125}^a$) > (${64}^a$) > $(125+64{)}^{a+b}$ > (${125}^b$)