# what is x if 3^(x-1),3^(x+1),5*3^x+1 are in arithmetic progresion ?

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You need to use the following property of consecutive members of an arithmetical sequence, such that:

`3^(x + 1) = (3^(x - 1) + 5*3^x + 1)/2`

You need to use the properties of exponents, such that:

`3*3^x = (3^x/3 + 5*3^x + 1)/2`

`2*3*3^x = 3^x/3 + 5*3^x + 1 => 6*3^x = 3^x/3 + 5*3^x + 1`

`3*6*3^x = 3^x + 15*3^x + 3 => 18*3^x = 16*3^x + 3`

You need to move the terms that contain `3^x` to the left side, such that:

`18*3^x - 16*3^x = 3 => 2*3^x = 3 => 3^x = 3/2`

Taking common logarithms both sides, yields:

`ln 3^x = ln(3/2)`

Using the power property of logarithms yields:

`x*ln 3 = ln(3/2) => x = ln(3/2)/ln 3`

**Hence, evaluating x, under the given conditions, yields **`x = ln(3/2)/ln 3.`