Having trouble with these
log |x – 1| = 3
log sub6 (log sub2 x) = 1
2007-02-27 11:25:28 · 2 answers · asked by Dylan 2 in Education & Reference ➔ Homework Help
First one looks like nondescript log base 10, so...
10^log|x-1| = 10^3
-(10^3) <= x - 1 <= 10^3
-(10^3) + 1 <= x <= (10^3) +1
Whenever you get rid of the absolute bracketing, you need to account for any and all possible combinations, thus the two inequalities.
Second one is much the same,
6^(log_6(log_2(x)) = (6^1)
log_2(x) = 6
2^log(x) = 2^6
x = 2^6
2007-02-27 11:37:46 · answer #1 · answered by Anonymous · 0⤊ 0⤋
log |x – 1| = 3
|x – 1| = e^3
x - 1 = e^3, 1 - x = e^3
x = 1 + e^3, x = 1 - e^3
x = 21.08554, -19.08554
log sub6 (log sub2 x) = 1
(log sub2 x) = 6^1 = 6
x = 2^6 = 64
2007-02-27 19:41:59 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋