3^(2x)-5(3^x)+4=0 , Find X?
2009-03-02 10:15 am
I know you have to use log but I'm not sure how to solve it, can someone do a step by step analysis of this? Thanks.
回答 (3)
2009-03-02 10:20 am
✔ 最佳答案
3^(2x) = (3^x)²Let u = 3^x
u²-5u+4 = 0
(u-4)(u-1) = 0
u = {1,4}
3^x = {1, 4}
x = {0, ln(4)/ln(3)}
2009-03-02 10:44 am
3^(2x) - 5(3^x) + 4 = 0
(3^x)^2 - 5(3^x) + 4 = 0
Change 3^x into y:
(3^x)^2 - 5(3^x) + 4 = 0
y^2 - 5y + 4 = 0
y^2 - y - 4y + 4 = 0
(y^2 - y) - (4y - 4) = 0
y(y - 1) - 4(y - 1) = 0
(y - 1)(y - 4) = 0
y - 1 = 0
y = 1
y - 4 = 0
y = 4
Change y into 3^x:
y = 1
3^x = 1
3^x = 3^0
x = 0
y = 4
3^x = 4
x = log_3(4)
∴ x = 0 , log_3(4)
(3^x)^2 - 5(3^x) + 4 = 0
Change 3^x into y:
(3^x)^2 - 5(3^x) + 4 = 0
y^2 - 5y + 4 = 0
y^2 - y - 4y + 4 = 0
(y^2 - y) - (4y - 4) = 0
y(y - 1) - 4(y - 1) = 0
(y - 1)(y - 4) = 0
y - 1 = 0
y = 1
y - 4 = 0
y = 4
Change y into 3^x:
y = 1
3^x = 1
3^x = 3^0
x = 0
y = 4
3^x = 4
x = log_3(4)
∴ x = 0 , log_3(4)
2009-03-02 10:23 am
Let y = 3^x
Then
3^(2x) - 5(3^x) + 4 = y^2 - 5y + 4 = 0
(y - 4)(y - 1) = 0
So y = 4 and y = 1
substitute y = 3^x back in to get
3^x = 4 and 3^x = 1
now take logs
3^x = 4
log(3^x) = log(4)
xlog(3) = log(4)
x = log(4)/log(3) = 2[log(2)/log(3)]
3^x = 1
log(3^x) = log(1) = 0
xlog(3) = 0
x = 0
Hope that helps :)
Then
3^(2x) - 5(3^x) + 4 = y^2 - 5y + 4 = 0
(y - 4)(y - 1) = 0
So y = 4 and y = 1
substitute y = 3^x back in to get
3^x = 4 and 3^x = 1
now take logs
3^x = 4
log(3^x) = log(4)
xlog(3) = log(4)
x = log(4)/log(3) = 2[log(2)/log(3)]
3^x = 1
log(3^x) = log(1) = 0
xlog(3) = 0
x = 0
Hope that helps :)
收錄日期: 2021-05-01 12:06:09
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