Solve: 2 log(c-6) = log (1/5) + log (20c-15)

Answer 1

2log(c - 6) = log (1/5) + log (20c - 15)

log(c - 6)² = log[(1/5)(20c - 15)]

log(c - 6)² = log(4c - 3)

(c - 6)² = (4c - 3)

c² - 12c + 36 = 4c - 3

c² - 16c + 39 = 0

(c - 13)(c - 3) = 0

c = 13 and c = 3

Plug them in to ensure both are valid (not extraneous).

2log(13 - 6) = log (1/5) + log (20*13 - 15) yes, this is true

2log(3 - 6) = log (1/5) + log (20*3 - 15) no, 3 - 6 gives -3 and -3 is not in the domain of log so this is not a real solution.

solution: c = 13