What is the answer to log_{3}((2x^2+5x+4)-7)=2 ?

The answer to log_{3}((2x^2+5x+4)-7)=2 is x= 3/2 ,x=-4

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log_{3}((2x^2+5x+4)-7)=2

lo g_{3} ((2 x^{2} + 5 x + 4) - 7) = 2

x = \frac{3}{2}, x = - 4

Decimal Notation

x = 1.5, x = - 4

Solution steps

lo g_{3} ((2 x^{2} + 5 x + 4) - 7) = 2

Apply log rules

2 x^{2} + 5 x - 3 = 9

Solve

2 x^{2} + 5 x - 3 = 9 : x = \frac{3}{2}, x = - 4

x = \frac{3}{2}, x = - 4

ln (2 x + 4) = ln (x + 3)

lo g_{3} (x + 12) = lo g_{3} (5 x)

lo g_{p} (- 1) = n

lo g_{5} (x - 3) + lo g_{5} (3 x + 1) = 3

lo g_{2 x} (216) = 3

What is the answer to log_{3}((2x^2+5x+4)-7)=2 ?
The answer to log_{3}((2x^2+5x+4)-7)=2 is x= 3/2 ,x=-4
Solve for x: log_{3}((2x^2+5x+4)-7)=2
The solution is x= 3/2 ,x=-4