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Mathematics 15 Online
OpenStudy (anonymous):

Solve the equation: 3^x+1=1/9

OpenStudy (anonymous):

you realize that 3^0 is 1 and 1/9 is 3^-2

OpenStudy (anonymous):

3^x+3^0=3^-2

OpenStudy (anonymous):

3^x+1=1/3^2 3^x+1=3^-2 x+1=-2 x=-3

OpenStudy (anonymous):

oh is it 3^(x+1)? no wonder, i was getting worried there

OpenStudy (anonymous):

Acually im not sure but two answers better than one right?

OpenStudy (anonymous):

nah, 3^0 doesn't work out

OpenStudy (anonymous):

that would be unsolvable

OpenStudy (anonymous):

Or it can be solved by log Log(3^x)+log(1)=log(3^-2) xlog3+log1=-2log3 Xlog3=-2log3-log1 X=(-2log3-log1)/log3

OpenStudy (anonymous):

don't think you can do that log(3^x+1) is not log(3^x)+log(1)

OpenStudy (anonymous):

you take the log of both sides, not individual expressions

OpenStudy (anonymous):

Hmmm... Will it does has something wrong but cant find it

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