Question 1004881
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Hello,

is your equation &nbsp;&nbsp;{{{(1/3)^(x+1)}}} = {{{1/729}}} ?

If so, then it is equivalent to 

{{{3^(x+1)}}} = {{{729}}} = {{{3^6}}},

and, &nbsp;hence, &nbsp;x + 1 = 6 &nbsp;and &nbsp;x = 6 - 1 = 5.

<U>Answer</U>. x = 5.
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