Advertisements
Advertisements
Question
Compute:
`(-3)^4-(root(4)(3))^0xx(-2)^5÷(64)^(2/3)`
Sum
Advertisements
Solution
`(-3)^4-(root(4)(3))^0xx(-2)^5÷(64)^(2/3)`
`=(-3xx-3xx-3xx-3)-1xx-2xx-2xx-2xx-2xx-2÷(2^6)^(2/3)`
Note : `(root(4)3)^0=1`
`=3^4+2^5÷2^(6xx2/3)`
`=3^4+2^5÷2^4=3^4+2^5/2^4`
`=3^4+2^(5-4)=3^4+2=3xx3xx3xx3+2`
`=81+2=83`
shaalaa.com
More About Exponents
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
Compute:
`1^8xx3^0xx5^3xx2^2`
Compute:
`(4^7)^2xx(4^-3)^4`
Compute:
`(-5)^4xx(-5)^6÷(-5)^9`
Evaluate:
`[(10^3)^0]^5`
Evaluate:
`9^0+9^-1-9^-2+9^(1/2)-9^(-1/2)`
Simplify:
`15"y"^8÷3"y"^3`
Simplify and express as positive indice:
`("a"^(-2)"b")^(1/2)xx("a""b"^-3)^(1/3)`
Evaluate:
`("a"^(2"n"+1)xx"a"^((2"n"+1)(2"n"-1)))/("a"^("n"(4"n"-1))xx("a"^2)^(2"n"+3)`
Prove that:
`1/(1+"x"^("a"-"b"))+1/(1+"x"^("b"-"a"))=1`
Simplify:
`("a"^(2"n"+3)."a"^((2"n"+1)("n"+2)))/(("a"^3)^(2"n"+1)."a"^("n"(2"n"+1)`
