What is the answer to log_{-2x-3}(3x^2+x-19)=2 ?

The answer to log_{-2x-3}(3x^2+x-19)=2 is x=-7,x=-4

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log_{-2x-3}(3x^2+x-19)=2

lo g_{- 2 x - 3} (3 x^{2} + x - 19) = 2

x = - 7, x = - 4

Solution steps

lo g_{- 2 x - 3} (3 x^{2} + x - 19) = 2

Apply log rules

\frac{ln ( 3 x ^{2} + x - 19 )}{ln ( - 2 x - 3 )} = 2

Multiply both sides by

ln (- 2 x - 3)

\frac{ln ( 3 x ^{2} + x - 19 )}{ln ( - 2 x - 3 )} ln (- 2 x - 3) = 2 ln (- 2 x - 3)

Simplify

ln (3 x^{2} + x - 19) = 2 ln (- 2 x - 3)

Apply log rules

3 x^{2} + x - 19 = (- 2 x - 3)^{2}

Solve

3 x^{2} + x - 19 = (- 2 x - 3)^{2} : x = - 7, x = - 4

x = - 7, x = - 4

Verify Solutions:

x = - 7

True

, x = - 4

True

The solutions are

x = - 7, x = - 4

lo g_{2} (x) + lo g_{2} (4 x) = 4

lo g_{x} (18) = 7.01

12 = lo g_{5} (x + 150) + 9

\frac{x ln ( x + 3 )}{x ^{2} + 1} = 0

lo g_{9} (a) = 12

What is the answer to log_{-2x-3}(3x^2+x-19)=2 ?
The answer to log_{-2x-3}(3x^2+x-19)=2 is x=-7,x=-4
Solve for x: log_{-2x-3}(3x^2+x-19)=2
The solution is x=-7,x=-4