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

solve for x:2log_(10)x-log_(x)(0.01)=5

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 xlog_(5)120+(x-3)-2*log_(5)(1-5^(x-3))=-log_(5)(0.2-5^(x-4))

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.

Solve for x.x^(log_(10)x+2)=10^(log_(10)x+2)

The least value of the expression 2(log)_(10)x-(log)_(x)(0.01), for x>1, is (1980,2M)(a)10(b)2(c)-0.01(d) None of

The least value of the expression 2(log)_(10)x-(log)_(x)(0.01)* for x>1 is (a) 10 (b) 2(c)-0.01(d)4

The least value of the expression 2(log)_(10)x-(log)_x(0. 01),forx >1, is a. 10 b. 2 c. -0.01 d. none of these