(1)/(10g_((x+6)/(3)))(log_(2)(x-1)/(x+2))>0

log_((pi+6)/(3))(log_(2)[(x-1)/(x+2)])>0

log_(7)((2x-6)/(2x-1))>0

log_(10)x-(1)/(2)log_(10)(x-(1)/(2))=log_(10)(x+(1)/(2))-(1)/(2)log_(10)(x+(1)/(8))

Consider the inequalities log_(5)(x-3)+(1)/(2)log_(5)3<(1)/(2)log_(5)(2x^(2)-6x+7) and log_(3)x+log_(sqrt(3))x+log_((1)/(3))x<6

(1)/(log_(3)(x+1))<(1)/(2log_(9)sqrt(x^(2)+6x+9))

The inverse of f(x)=(10^(x)-10^(-x))/(10^(x)+10^(-x)) is A). (1)/(2)log_(10)((1+x)/(1-x)) , B). log_(10)(2-x) , C). (1)/(2)log_(10)(2-1) , D). (1)/(4)log_(10)((2x)/(2-x))

Solve for x, (a) (log_(10)(x-3))/(log_(10)(x^(2)-21))=(1)/(2),(b)log(log x)+log(log x^(3)-2)=0; where base of log is 10 everywhere.

(6)/(5)a^((log_(a)x)(log_(10)a)(log_(a)5))-3^(log_(10)((x)/(10)))=9^(log_(100)x+log_(4)2) (where a gt 0, a ne 1) , then log_(3)x=alpha +beta, alpha is integer, beta in [0, 1) , then alpha=

(6)/(5)a^((log_(a)x)(log_(10)a)(log_(a)5))-3^(log_(10)((x)/(10)))=9^(log_(100)x+log_(4)2)("where "a gt 0, a ne 1) , then log_(3)x=alpha +beta, alpha is integer, beta in [0, 1) , then alpha=

(6)/(5)a^((log_(a)x)(log_(10)a)(log_(a)5))-3^(log_(10)((x)/(10)))=9^(log_(100)x+log_(4)2)("where "a gt 0, a ne 1) , then log_(3)x=alpha +beta, alpha is integer, beta in [0, 1) , then alpha=