Maths help! How can i solve n in the following question? 3^(2n-4) = 1? Thanks!?
回答 (4)
2016-07-06 7:53 am
3^(2n - 4) = 1
3^(2n - 4) = 3^0
2n - 4 = 0
2n = 4
n = 2
3^(2n - 4) = 3^0
2n - 4 = 0
2n = 4
n = 2
2016-07-06 7:53 am
3^(2n-4) = 1
ln(3^(2n-4)) = ln(1)
(2n-4)*ln(3) = ln(1)
2n-4 = ln(1)/ln(3)
n = (ln(1)/ln(3)+4)/2
n = 2
ln(3^(2n-4)) = ln(1)
(2n-4)*ln(3) = ln(1)
2n-4 = ln(1)/ln(3)
n = (ln(1)/ln(3)+4)/2
n = 2
2016-07-06 8:12 am
3^(2n - 4) = 1
9^(n - 2) = 1
Any number to the power 0 is 1.
n - 2 = 0
n = 2
9^(n - 2) = 1
Any number to the power 0 is 1.
n - 2 = 0
n = 2
2016-07-06 8:07 am
use log to bring down the power or just make the bases the same and equate the powers. come on it's simple.
3^(2n-4) = 1
log3^(2n-4) = log1 --> take log base 10 to both sides
(2n-4)log3 = 0
2n - 4 = 0
n = 2
3^(2n-4) = 1
3^(2n-4) = 3^0 --> anything to the power of zero is 1 except 0
bases are the same therefore equate power
2n - 4 = 0
n = 2
3^(2n-4) = 1
log3^(2n-4) = log1 --> take log base 10 to both sides
(2n-4)log3 = 0
2n - 4 = 0
n = 2
3^(2n-4) = 1
3^(2n-4) = 3^0 --> anything to the power of zero is 1 except 0
bases are the same therefore equate power
2n - 4 = 0
n = 2
收錄日期: 2021-04-20 16:28:17
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