Question 379732
  <pre><font size = 3 color = "indigo"><b>
Hi,
2logX-log(8x-160)-1=0
*[tex \large\ \ alog_bx = log_b(x)^a ]
*[tex \large\ \ log_bx - log_by = log_b(x/y) ]
log(x^2/8(x - 20)=1
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ log_b(x) \ = \ y \ \ \Rightarrow\ \ b^y = x]
10^1  = (x^2/8(x - 20)
10*8(x-20) = x^2
80x - 1600 = x^2
x^2 -80x + 1600 = 0
(x-40)(x-40)=0
x = 40