solve for x: `2log_10x-log_x(0.01)=5`

Solve for x: log_(a)x+log_(x)a=2

For x > 1 , the minimum value of 2 log_10(x)-log_x(0.01) is

For x > 1 , the minimum value of 2 log_10(x)-log_x(0.01) is

For x > 1 , the minimum value of 2 log_10(x)-log_x(0.01) is

For x>1, the minimum value of 2log_(10)(x)-log_(x)(0.01) is

Solve for x: a) (log_(10)(x-3))/(log_(10)(x^(2)-21)) = 1/2 b) log(log x)+log(logx^(3)-2)= 0, where base of log is 10. c) log_(x)2. log_(2x)2 = log_(4x)2 d) 5^(logx)+5x^(log5)=3(a gt 0), where base of log is 3. e) If 9^(1+logx)-3^(1+logx)-210=0 , where base of log is 3.

For x>1, show that: 2log_(10)x-log_(x)0.01>=4

Solve for x: a) log_(x)2. log_(2x)2 = log_(4x)2 b) 5^(logx)+5x^(log5)=3(a gt 0), where base of log is 3.

Solve for x, if : log_(x)49 - log_(x)7 + "log"_(x)(1)/(343) + 2 = 0 .