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=(-5+sqrt(157457))/4 ,x=(-5-sqrt(157457))/4

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

The solution is x=(-5+sqrt(157457))/4 ,x=(-5-sqrt(157457))/4

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

x = \frac{- 5 + 157457}{4}, x = \frac{- 5 - 157457}{4}

Decimal Notation

x = 97.95212 \dots, x = - 100.45212 \dots

Solution steps

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

Move

7

to the right side

lo g_{3} (2 x^{2} + 5 x + 4) = 9

Apply log rules

2 x^{2} + 5 x + 4 = 19683

Solve

2 x^{2} + 5 x + 4 = 19683 : x = \frac{- 5 + 157457}{4}, x = \frac{- 5 - 157457}{4}

x = \frac{- 5 + 157457}{4}, x = \frac{- 5 - 157457}{4}

Verify Solutions:

x = \frac{- 5 + 157457}{4}

True

, x = \frac{- 5 - 157457}{4}

True

The solutions are

x = \frac{- 5 + 157457}{4}, x = \frac{- 5 - 157457}{4}

\frac{lo g _{10} ( 6 - x ^{2} )}{lo g _{10} ( 3 x - 4 )} = 2

lo g_{4} (x + 12) - lo g_{4} (5 - x) = 2

lo g_{2} (x + 7) = 5

lo g_{b} (3) = \frac{1}{2}

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

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=(-5+sqrt(157457))/4 ,x=(-5-sqrt(157457))/4
Solve for x: log_{3}(2x^2+5x+4)-7=2
The solution is x=(-5+sqrt(157457))/4 ,x=(-5-sqrt(157457))/4