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OpenStudy (anonymous):
We can take the cube root of both sides to get
(3x)^2 = 3^(-2)
Now we can "take the square root", but must keep in mind that there is a positive and a negative answer, so that
\[3x = \pm(3^{-1}) = \pm{1 \over 3}\]and hence
\[x = \pm{1 \over 9}\]I'm assuming x is only allowed to be real. There are four additional complex solutions to this (which would come out at the point that I took the cube root), but I won't bother typing it out unless you want/need to know what they are.
OpenStudy (anonymous):
the answer is 1/9 ?
OpenStudy (mertsj):
Write both sides with the exponent of 6
OpenStudy (anonymous):
positive or negative 1/9
Because a negative to an even power is positive.
OpenStudy (mertsj):
\[(3x)^6=(\frac{1}{3})^6\]
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OpenStudy (anonymous):
no it (3^x)^6=3^-6
OpenStudy (mertsj):
Since the expressions are equal and the exponents are the same, the bases must also be the same so write:
\[3x=\frac{1}{3}\]
OpenStudy (mertsj):
Solve that.
OpenStudy (mertsj):
\[(3^x)^6=3^{6x}\]
OpenStudy (mertsj):
So:
\[3^{6x}=3^{-6}\]
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