Solve:
log3(x-4)=2-log3(x+4)
The 3 next to the logs are both subscripts.
Math Help!?
2017-12-08 9:29 pm
回答 (5)
2017-12-08 11:17 pm
✔ 最佳答案
Some general rules:log(x) + log(y) = log(xy)
a^logₐ(x) = x
log₃(x-4) = 2-log₃(x+4)
log₃(x-4) + log₃(x+4) = 2
log₃((x-4)(x+4)) = 2
log₃(x²-16) = 2
x²-16 = 3²
x² = 25
x = ±√25 = ±5
The argument of a logarithm must be positive, so -5 is an extraneous solution.
x = 5
2017-12-08 10:08 pm
log₃(x - 4) = 2 - log₃(x + 4)
log₃(x - 4) + log₃(x + 4) = 2log₃(3)
log₃(x - 4)(x + 4) = log₃(3²)
(x - 4)(x + 4) = 3²
x² - 16 = 9
x² = 25
x = 5 or x = -5 [rejected, for log₃(-5 - 4) = log₃(-9) is undefined]
Hence, x = 5
log₃(x - 4) + log₃(x + 4) = 2log₃(3)
log₃(x - 4)(x + 4) = log₃(3²)
(x - 4)(x + 4) = 3²
x² - 16 = 9
x² = 25
x = 5 or x = -5 [rejected, for log₃(-5 - 4) = log₃(-9) is undefined]
Hence, x = 5
2017-12-08 9:35 pm
log(base 3)(x - 4) = 2 - log(base 3)(x + 4)
log(base 3)(x - 4) + log(base 3)(x + 4) = 2
log(base 3)((x - 4)(x + 4)) = 2
log(base 3)(x^2 - 16) = 2
3^2 = x^2 - 16
x^2 = 25
x = 5 or x = -5
Discard x = -5 because it leads to imaginary numbers, thus a wrong solution, in the original equation, leaving just
x = 5
log(base 3)(x - 4) + log(base 3)(x + 4) = 2
log(base 3)((x - 4)(x + 4)) = 2
log(base 3)(x^2 - 16) = 2
3^2 = x^2 - 16
x^2 = 25
x = 5 or x = -5
Discard x = -5 because it leads to imaginary numbers, thus a wrong solution, in the original equation, leaving just
x = 5
2017-12-08 10:21 pm
Let log be log to base 3
log [ ( x - 4 ) ( x + 4 ) ] = 2
( x - 4 ) ( x + 4 ) = 9
x² = 25
x = 5
log [ ( x - 4 ) ( x + 4 ) ] = 2
( x - 4 ) ( x + 4 ) = 9
x² = 25
x = 5
2017-12-08 9:53 pm
log₃(x-4) = 2 - log₃(x+4)
log₃(x - 4) + log₃(x + 4) = 2
log₃[(x - 4)(x + 4)] = 2
log₃[x^2 - 16] = 2
x^2 - 16 = 3^2
x^2 = 16 + 9
x^2 = 25
x = ± 5
log₃(x - 4) + log₃(x + 4) = 2
log₃[(x - 4)(x + 4)] = 2
log₃[x^2 - 16] = 2
x^2 - 16 = 3^2
x^2 = 16 + 9
x^2 = 25
x = ± 5
收錄日期: 2021-04-18 18:04:05
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