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log(3)a-log(x)a=log(x/3)a 3,x and x/3 are in the base

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log(3)a-log(x)a=log(x/3)a

3,x and x/3 are in the base

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You need to use the following logarithmic identity such that:

`log_3 a = 1/(log_a 3)`

`log_x a = 1/(log_a x)`

`log_(x/3) a = 1/(log_a (x/3))`

`1/(log_a 3)- 1/(log_a x) = 1/(log_a (x/3))`

`(log_a x - log_a 3)(log_a (x/3)) = (log_a 3)(log_a x)`

`(log_a x)^2 - 2(log_a 3)(log_a x) + (log_a 3)^2 = (log_a 3)(log_a x)`

`(log_a x)^2 - 3(log_a 3)(log_a x) + (log_a 3)^2 = 0`

Coming up with the substitution `log_a x = y`  yields:

`y^2 - 3(log_a 3)*y + (log_a 3)^2 = 0`

`y_(1,2) = (3(log_a 3) +- sqrt(9(log_a 3)^2 - 4(log_a 3)^2))/2 `

`y_(1,2) = (3(log_a 3) +- (log_a 3)sqrt5)/2`

`y_(1,2) = (log_a 3)(3+-sqrt5)/2`

You need to solve for x the equations `log_a x = (log_a 3)(3+-sqrt5)/2:` `log_a x = (log_a 3)(3+sqrt5)/2 => x = a^((log_a 3)(3+sqrt5)/2)`

`log_a x = (log_a 3)(3-sqrt5)/2 => x = a^((log_a 3)(3-sqrt5)/2)`

Hence, evaluating the solutions to the equation yields `x = a^((log_a 3)(3+sqrt5)/2)`  and `x = a^((log_a 3)(3-sqrt5)/2).`

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