# solve for x,y `x*log_3 y = 4` `x^2*log_3 y = 8`

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### 2 Answers

The equations to be solved for x and y are:

`x*log_3 y = 4` ...(1)

`x^2*log_3 y = 8` ...(2)

(2)/(1)

=> `(x^2*log_3 y)/(x*log_3 y)` = `8/4` = 2

=> x = 2

Substitute in (1)

=> `2*log_3 y = 4`

=> `log_3 y = 2`

=> `y = 3^2 = 9`

**The solution of the system of equations is x = 2 and y = 9**

You need to solve the system of simultaneous equations such that:

`x*log_3 y = 4`

`x^2*log_3 y = 8`

You should express `log_3 y` in terms of x such that:

`log_3 y = 4/x`

Substituting `4/x` for `log_3 y` in the equation `x^2*log_3 y = 8` yields:

`x^2*(4/x)= 8`

You need to subtract 8 both sides such that:

`4x^2/x - 8 = 0 =gt (4x^2 - 8x) = 0`

You need to factor out 4x such that:

`4x(x - 2) = 0 =gt 4x = 0 =gt x = 0`

`x - 2 = 0 =gt x = 2`

You need to substitute 0 for x in equation `x*log_3 y = 4` , hence `0*log_3 y = 4 =gt 0 != 4`

Notice that if x=0, the equation `x*log_3 y = 4` does not hold, hence 0 is not a solution to equation.

You need to substitute 2 for x in equation `x*log_3 y = 4` , hence `2*log_3 y = 4 =gt log_3 y = 2 =gt y = 3^2 =gt y = 9` .

**Hence, the solution to simultaneous equation is (2;9).**