Question Number 104199 by hardylanes last updated on 20/Jul/20

solve for real values of x the equation  4(3^(2x+1) )+17(3^x )=7.  if m and n are positive real numbers other  than 1, show that the log_n m+log_(1/m) n=0

Answered by bemath last updated on 20/Jul/20

12.3^x +17.3^x −7=0  3^x = ((−17+(√(289+48.7)))/(24))  3^x  = ((−17+25)/(24)) = (1/3)  ∴ x = −1

Answered by OlafThorendsen last updated on 20/Jul/20

4(3^(2x+1) )+17(3^x ) = 7  12(3^x )^2 +17(3^x )−7 = 0  u = 3^x   12u^2 +17u−7 = 0  Δ = 17^2 −4×12×(−7) = 625 = 25^2   u = ((−17±25)/(24))  u = 3^x  ⇒ u>0  then u = ((−17+25)/(24)) = (1/3)  u = 3^x  = (1/3) ⇒ x = −1  I think your second  question is false  (it′s true only for m = n  or m = (1/n)) but :  log_n m+log_(1/n) m =  ((lnm)/(lnn))+((lnm)/(ln(1/n))) =   ((lnm)/(lnn))−((lnm)/(lnn)) = 0