Determine x given that `9^x=(1/3)^(1-x)`

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The value of x has to be determined for which `9^x=(1/3)^(1-x)`

`9^x=(1/3)^(1-x)`

=> `(3^2)^x = 3^(x -1)`

=> `3^(2x) = 3^(x - 1)`

As the base is the same equate the exponent

=> `2x = x - 1`

=> `x = -1`

The solution of the equation `9^x=(1/3)^(1-x)` is x = -1

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To do this we need the following.

`log(a^n) = nloga`

`log(a/b) = loga-logb`

 

`9^x = (1/3)^(1-x)`

Get log on both sides.

`log(9^x) = log((1/3)^(1-x))`

`xlog9 = (1-x)log(1/3)`

`xlog9 = (1-x)(log1-log3)`

`xlog9 = (1-x)(0-log3)`

`xlog9 = -log3+xlog3`

`xlog3^2 = xlog3-log3`

`2xlog3 = xlog3-log3`

`xlog3 = -log3`

`x = -1`

 

So the answer is x = -1

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