Evaluate for x :
`( log_10(5) + log_10( x^2 + 1) ) /(log_10(x -2)) = 2`

log_(10)^(2) x + log_(10) x^(2) = log_(10)^(2) 2 - 1

(log_(10)x)^(2)+log_(10)x^(2)=(log_(10)2)^(2)-1

Solve for x. log_(0.1) sin 2x +log_10 cos x = log_10 7 .

Solve x+log_10(2^x+1)= log_10 6+xlog_10 5 .

Given that log_10 2 =x, log_10 3 =y then log_10 1.2=

Solve for x:x+(log)_(10)(1+2^(x))=x log_(10)5+log_(10)6

Solve the following equations for x and y: log_10x+log_10(x)^(1/2)+log_10(x)^(1/4)+….=y (1+3+5+…+(2y-1))/(4+7+10+..+(3y+1))=20/(7log_10x)

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.

Solution set of the in equality log_(10^(2)) x-3(log_(10)x)( log_(10)(x-2))+2 log_(10^(2))(x-2) lt 0 , is :