# Explain how solve indices equation `2^((2x-1)/(x-1))+2^((3x-2)/(x-1))=24` ? is a special technique ?

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You should come up with the following substitution such that:

`y = 1/(x-1) => x-1 = 1/y => x = 1/y + 1 => x = (1+y)/y => 2x = 2(1+y)/y => 2x - 1 = 2(1+y)/y - 1 => 2x-1 = (2+y)/y`

`3x = 3(1+y)/y => 3x - 2 = 3(1+y)/y - 2 => 3x - 2 = (3+y)/y`

`2^((2x-1)/(x-1)) = 2^(2x-1)*y => 2^((2x-1)/(x-1)) =2^(((2+y)/y)*y) => 2^((2x-1)/(x-1)) = 2^(2+y)`

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

Hence, substituting `2^(2+y)` for `2^((2x-1)/(x-1))` and `2^(3+y)` for `2^((3x-2)/(x-1))` yields:

`2^(2+y) + 2^(3+y) = 24`

You need to use the following identity: `a^(x+y) = a^x*a^y` such that:

`2^2*2^y + 2^3*2^y = 24 => 4*2^y+8*2^y = 24 => 12*2^y = 24`

`2^y = 24/12 => 2^y = 2 => y = 1`

You need to solve for x the equation `x = (1+y)/y` , hence, substituting 1 for y in this equation yields:

`x = (1+1)/1 => x = 2`

**Hence, evaluating the solution to the given equation yields x = 2.**