Solve for x: log base 3(t-2) + log base 3 (t+4)=3?
回答 (2)
2009-06-23 1:18 pm
✔ 最佳答案
log base3 (t^2+2t-8)=3t^2+2t-8=27
t^2+2t-35=0
t=[-1+&-(1+35)^1/2]=5
t=5 only is accepted
2009-06-23 3:01 pm
log_3(t - 2) + log_3(t + 4) = 3
log_3[(t - 2)(t + 4)] = 3
t*t + t*4 - 2*t - 2*4 = 3^3
t^2 + 4t - 2t - 8 - 27 = 0
t^2 + 2t - 35 = 0
t^2 + 7t - 5t - 35 = 0
(t^2 + 7t) - (5t + 35) = 0
t(t + 7) - 5(t + 7) = 0
(t + 7)(t - 5) = 0
t + 7 = 0
t = -7
t - 5 = 0
t = 5
(negative number is not acceptable)
â´ t = 5
log_3[(t - 2)(t + 4)] = 3
t*t + t*4 - 2*t - 2*4 = 3^3
t^2 + 4t - 2t - 8 - 27 = 0
t^2 + 2t - 35 = 0
t^2 + 7t - 5t - 35 = 0
(t^2 + 7t) - (5t + 35) = 0
t(t + 7) - 5(t + 7) = 0
(t + 7)(t - 5) = 0
t + 7 = 0
t = -7
t - 5 = 0
t = 5
(negative number is not acceptable)
â´ t = 5
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