Question 394657
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Hi
log(5)-log(2x)= -4
*[tex \LARGE\ \ log_bx - log_by = log_b(x/y) ]
log(5/2x)= -4
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ log_b(x) \ = \ y \ \ \Rightarrow\ \ b^y = x]
10^-4 = 5/2x
{{{ 1/10000 = 5/(2x )}}}   |cross multiplying
 2x = 50000
  x = 25000