How do you solve #log(x+54) - log(x+2) = 2 - log(x)#?

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Answer 1 · 2016-07-01T09:46:04

#28+2sqrt 246#=59.36877#, nearly

Use, #log ab= loga +logb and log(a/b)=loga -logb#

Rearranging.

#log((x(x+54))/(x+2))=2#

Inversely,

#((x(x+54))/(x+2))=10^2=100#.

Cross-multiply, rearrange and solve for positive root..

#x^2-56 x- 200=0#

#x=28+2sqrt 246#=59.36877#, nearly