Question 93454
You are on the right track with:
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{{{9^y = 1/3}}}
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Now you can make a couple of changes. First, recognize that {{{1/3}}} is equivalent to
{{{3^-1}}}.
Substitute this into the equation to get:
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{{{9^y = 3^(-1)}}}
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Now notice that 9 is equal to {{{3^2}}}. So replace the 9 with {{{3^2}}} and you get:
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{{{(3^2)^y = 3^(-1)}}}
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by the power rule of exponents you can multiply the exponent y by the exponent 2 to make the
left side of the equation change as indicated:
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{{{3^(2y) = 3^(-1)}}}
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Notice that the bases of the two sides are both 3. Therefore, for the sides to be equal, the
two exponents must also be equal. When you set the two exponents equal you get:
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{{{2y = -1}}}
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Solve for y by dividing both sides by 2 to get:
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{{{y = (-1)/2 = -1/2}}}
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That's the answer for y. 
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Check it out by substituting {{{-1/2}}} for y in the original problem to get ...
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{{{log(9,(1/3)) = -1/2}}}
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When you convert this to exponential form you get:
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{{{9^(-1/2) = 1/3}}}
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But a negative exponent becomes a positive exponent if you put it (along with its base) into
the denominator of a fraction. So {{{9^(-1/2)}}} becomes {{{1/(9^(1/2))}}} and the 
equation is then:
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{{{1/9^(1/2) = 1/3}}}
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Then recognize that the exponent {{{1/2}}} tells you to take the square root of its base. So
{{{9^(1/2)}}} means to take the square root of 9 and the square root of 9 is 3.
So you can 
replace {{{9^(1/2)}}} by 3 and the equation is finally reduced to:
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{{{ 1/3 = 1/3}}}
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Since this is true, you know that {{{y = -1/2}}} is the correct answer.
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Hope this helps you to understand where you could go with the problem.  You had a good start
at it by correctly changing the logarithmic form to the exponential form.