# what is x,y?log base0.1 x+log base2y=2 5xy=1

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### 1 Answer

You need to use the formula change the base of logarithm such that:

`log_a b = (log_c b)/(log_c a)`

`log_0.1 x = (log_2 x)/(log_2 10^(-1))`

`(log_2 x)/(log_2 10^(-1)) + log_2 y = 2`

You need to consider the equation `5xy=1` such that:

`xy = 1/5 =gt log_2 (xy) = log_2 (1/5)`

You need to conver the logarithm of product in the sum of logarithms such that:

`log_2 x + log_2 y = -log_2 5 =gt log_2 y =-log_2 5 --log_2 x`

`(log_2 x)/(log_2 10^(-1)) -log_2 5 - log_2 x = 2`

`log_2 x + log_2 10*log_2 5 + log_2 10*log_2 x = 2`

You need to factor out `log_2 x` such that:

`log_2 x(1 + log_2 10) = 2 - log_2 10*log_2 5`

`log_2 x =(2 - log_2 10*log_2 5)/(log_2 20)`

`x = 2^((2 - log_2 10*log_2 5)/(log_2 20))`

You need to substitute `(2 - log_2 10*log_2 5)/(log_2 20)` for `log_2 x ` in `log_2 y = -log_2 5 -log_2 x` such that:

`log_2 y = -log_2 5 - (2 - log_2 10*log_2 5)/(log_2 20)`

`y = 2^(-log_2 5 - (2 - log_2 10*log_2 5)/(log_2 20))`

**Hence, the solution to the simultaneous equations is `x = 2^((2 - log_2 10*log_2 5)/(log_2 20))` and `y = 2^(-log_2 5 - (2 - log_2 10*log_2 5)/(log_2 20)).` **