How would you solve : 3^(x+1) = -3^x + 8
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How would you solve : 3^(x+1) = -3^x + 8

[From: ] [author: ] [Date: 11-09-06] [Hit: ]
......
3^x+1 +3^x=8
3^x(3+1)=8
3^x(4)=8
3^x=2
xlog3=log2
x=log2/log3

~David

-
3^(x + 1) = - 3^x + 8
3^(x + 1) + 3^x = 8
3(3^x) + 3^x = 8
4(3^x) = 8
3^x = 2
xln3 = ln 2
x = ln2/ln3
x = .69/1.1
x = .62

3^(1.62) = - 3^(.62) + 8
6 = - 2 + 8
6 = 6
1
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