What is the answer to log_{-2x-8}(2x^2+6x-20)=2 ?

The answer to log_{-2x-8}(2x^2+6x-20)=2 is x=-7,x=-6

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log_{-2x-8}(2x^2+6x-20)=2

lo g_{- 2 x - 8} (2 x^{2} + 6 x - 20) = 2

x = - 7, x = - 6

Solution steps

lo g_{- 2 x - 8} (2 x^{2} + 6 x - 20) = 2

Apply log rules

\frac{ln ( 2 x ^{2} + 6 x - 20 )}{ln ( - 2 x - 8 )} = 2

Multiply both sides by

ln (- 2 x - 8)

\frac{ln ( 2 x ^{2} + 6 x - 20 )}{ln ( - 2 x - 8 )} ln (- 2 x - 8) = 2 ln (- 2 x - 8)

Simplify

ln (2 x^{2} + 6 x - 20) = 2 ln (- 2 x - 8)

Apply log rules

2 x^{2} + 6 x - 20 = (- 2 x - 8)^{2}

Solve

2 x^{2} + 6 x - 20 = (- 2 x - 8)^{2} : x = - 7, x = - 6

x = - 7, x = - 6

Verify Solutions:

x = - 7

True

, x = - 6

True

The solutions are

x = - 7, x = - 6

v, ln (v) = - 2 t + ln (20)

lo g_{2 x - 5} (2 x^{2} - 2 x - 15) = 2

lo g_{12} (1 + \frac{r}{100}) = 3

lo g_{5} (x) - lo g_{5} (x - 1) = 1

lo g_{6} (x + 4) = 1 - lo g_{6} (x + 3)

What is the answer to log_{-2x-8}(2x^2+6x-20)=2 ?
The answer to log_{-2x-8}(2x^2+6x-20)=2 is x=-7,x=-6
Solve for x: log_{-2x-8}(2x^2+6x-20)=2
The solution is x=-7,x=-6