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December 19, 2014

December 19, 2014

Posted by **Semja** on Tuesday, December 13, 2011 at 4:47pm.

3^(2x+1)-16(3^x)+5=0

- Math - Exponentials -
**bonita**, Tuesday, December 13, 2011 at 5:07pmDistribute

- Math - Exponentials -
**Semja**, Tuesday, December 13, 2011 at 5:16pmEventually 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

- Math - Exponentials -
**Reiny**, Tuesday, December 13, 2011 at 6:18pmI 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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