# Solve the equation: `log_30(x^2+ x) = 1`

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Solve `log_(30)(x^2+x)=1` :

`log_(b)a=c==> b^c=a` (e.g. `log_2 8=3 ==> 2^3=8` )

so we can rewrite as:

`x^2+x=30^1`

`x^2+x-30=0`

`(x+6)(x-5)=0`

x=-6 or 5

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The solutions are x=-6 or x=5

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Check: `log_(30)((-6)^6-6)=log_(30)30=1`

and `log_(30)(5^2+5)=log_(30)30=1`

Since `log_(30)(x^2+x)=1` That means that `x^2+x=30^1=30` So now you have a quadratic equation that can be solved for 0 and factored. `x^2+x-30=0` We factor that and get `(x+6)(x-5)=0` so x=5, -6