`3^(log_(3)7)+7^(log_(8)(1/8))+log_(0.3)3`

If 3^(log_(3)7)+7^(log_(8)((1)/(8)))+log_(0.bar(3))(3)=(a)/(b) , a ,b in N HCF(a , b)=1 then

7^(log_(3)5)+3^(log_(5)7)-5^(log_(3)7)-7^(log_(5)3)

3(log_(7)x+log_(x)7)=10

Calculate 7^(log g_(3)5) + 3 ^(log_(5)7 ) - 5 ^(log_(3)7) - 7 ^(log_(5)3)

log_(3)(5+x)+log_(8)8=2^(2)

For N>1, then product (1)/(log_(2)N)*(1)/(log_(N)8)*(1)/(log_(32)N)*(1)/(log_(N)128) simplifies to (a) (3)/(7) (b) (3)/(7ln2) (c) (3)/(5ln2)(d)(5)/(21)

Find the value of (log_(3)4)(log_(4)5)(log_(5)6)(log_(6)7)(log_(7)8)(log_(8)9)

log_((3)/(4))log_(8)(x^(2)+7)+log_((1)/(2))log_((1)/(4))(x^(2)+7)^(-1)=-2