Posted by Semja on Tuesday, December 13, 2011 at 4:47pm.
Distribute
Eventually answered my own question. No idea what distributing has to do with anything though...
I simplified 3^(2x+1) to ((3^x)^2)*3^x then substituted a variable for 3^x and solved the corresponding quadratic. Seems to have worked
I think you were on the right track, except
3^(2x+1) = 3(3^x)(3^x) = 3(3^x)^2
so let y = 3^x to change your equation to
3y^2 - 16y + 5 = 0
(y - 5)(3y - 1) = 0
y = 5 or y = 1/3
so 3^x = 5 or 3^x = 1/3
This cannot be solved WITHOUT a calculator
3^x = 5
log 3^x = log 5
x log3 = log 5
x = log5/log3 = appr. 1.465
but 3^x = 1/3
3^x = 3^-1
x = -1
so x = -1 or x = appr. 1.465
both answers verify in the original
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