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

Solve for x : log_(10)x = -2 .

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

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 .

For xgt1 , show that: 2log_10x-log_x 0.01ge4

Solve for x : (i) log_(10) (x - 10) = 1 (ii) log (x^(2) - 21) = 2 (iii) log(x - 2) + log(x + 2) = log 5 (iv) log(x + 5) + log(x - 5) = 4 log 2 + 2 log 3

The least value of the expression 2(log)_(10)x-(log)_x(0.01)dot for x >1 is (a) 10 (b) 2 (c) -0. 01 (d) 4

Solve for x : 3^(log x)-2^(log x) =2^(log x+1)-3^(log x-1)

Solve for x :(log)_5 120+(x-3)-2.(log)_5(1-5^(x-3))=-(log)_5(0. 2-5^(x-4))dot

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