Can you please help me solve for x
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Can you please help me solve for x

[From: ] [author: ] [Date: 12-07-13] [Hit: ]
There is no value suitable for your equation among real numbers.Divide both side by 9.And the exponents must be equal so x+1=-2.x=-3.-By 3^x+1 do you mean 3^(x+1) or (3^x)+1?In the former care x = -3 in the later case you have to use log.......
27(3^x+1)=3

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DIVIDE BY 3 ON BOTH SIDES
9(3^x+1)=1
3*3*(3^x)+9=1
3^(2+x)+9=1
3^(x+2)= -- 8
take log
(x+2)log3=log (- 8)
x+2=log -3/log3
x=log -3/log3-2
x has no real value since log of negative number doesnt exist

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27(3^x+1)=3
9(3^x+1)=1
9x3^x+9=1
(3^2)x(3^x)=-8
3^(x+2)=-8
here you know none of powers of 3 equals minus value. As well as
lg[3^(x+2)]=lg[-8]
lg negative values are indetermined.
There is no value suitable for your equation among real numbers.

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Divide both side by 3



9(3^x+1)=1

Divide both side by 9.


3^x+1= 1/9

1/9 is the same thing as 3^-2

And the exponents must be equal so x+1=-2.
x=-3.

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By 3^x+1 do you mean 3^(x+1) or (3^x)+1?
In the former care x = -3 in the later case you have to use log.

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27(3^x+1)=3
(3^x+1)=3/27
3^x=(3/27)-1
ln(3^x)=ln((3/27)-1)
x ln3=ln((3/27)-1)
x = (ln((3/27)-1)) / ln 3

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3^(x + 1) = 3^(-2)
x + 1 = - 2
x = - 3
1
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