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

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embizze's profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

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=-6 or 5


The solutions are x=-6 or x=5


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

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

tjbrewer's profile pic

tjbrewer | Elementary School Teacher | (Level 2) Associate Educator

Posted on

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

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