laws of rational exponent (v)` (ab)^p` =` a^p b^p` (vi)`(a/b)^p` =` a^p/b^p` (vii) `a^(p/q) = (a^p)^(1/q) = (a^q)^(1/p)`

laws of Rational exponents are same as real exponents.(i) a^(p)xx a^(q)=a^(p+q)( ii) (a^(p))/(a^(q))=a^(p-q) (iii) (a^(p))^(q)=a^(pq)( iv )a^(-q)=(1)/(a^(q))

Prove that a^(p)b^(q)<((ap+bq)/(p+q))^(p+q)

If in a g.P. { t_(n)) it is given that t_(p+q) =a and t_(p-q) = b then : t_(p) = (A) (ab)^(1/2) (B) (ab)^(1/3) (C) (ab)^(1/4) (D) none of these

Prove that a^p b^ q <((a p+b q)/(p+q))^(p+q)dot

Prove that a^p b^ q <((a p+b q)/(p+q))^(p+q)dot

Prove that a^p b^ q <((a p+b q)/(p+q))^(p+q)dot

((p+1/q)^p.(p-1/q)^q)/((q+1/p)^p.(q-1/p)^q)=(p/q)^x then x=

(2) If a=x^(q+r)*y^(p) , b=x^(r+p)*y^(q) , c=x^(p+q)y^(r) , show that a^(q-r)b^(r-p)c^(p-q)=1

The logically equivalent proposition of phArrq is (p^^q)vv(p^^q) b. (p=>q)^^(p=>q) c. (p^^q)vv(p=>q) d. (p^^q)(pvvq)

~[(~p) ^^q] is logically equivalent to a) ~ (p vv q) b) ~ [p ^^ (~q)] c) p ^^ (~q) d) p vv (~q)