What is the answer to 1=log_{10}(2x^2+3x-22) ?

The answer to 1=log_{10}(2x^2+3x-22) is x=(-3+sqrt(265))/4 ,x=(-3-sqrt(265))/4

Solve for x: 1=log_{10}(2x^2+3x-22)

The solution is x=(-3+sqrt(265))/4 ,x=(-3-sqrt(265))/4

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1 = lo g_{10} (2 x^{2} + 3 x - 22)

x = \frac{- 3 + 265}{4}, x = \frac{- 3 - 265}{4}

Decimal Notation

x = 3.31970 \dots, x = - 4.81970 \dots

Solution steps

1 = lo g_{10} (2 x^{2} + 3 x - 22)

Apply log rules

10 = 2 x^{2} + 3 x - 22

Solve

10 = 2 x^{2} + 3 x - 22 : x = \frac{- 3 + 265}{4}, x = \frac{- 3 - 265}{4}

x = \frac{- 3 + 265}{4}, x = \frac{- 3 - 265}{4}

lo g_{x - 8} (x^{2} - 16) = 2

lo g_{1} (x) = 1.755

lo g_{10} (3 x - 3) = 3

y, x = 10 lo g_{2} (y)

2 = lo g_{x} (3^{2})

What is the answer to 1=log_{10}(2x^2+3x-22) ?
The answer to 1=log_{10}(2x^2+3x-22) is x=(-3+sqrt(265))/4 ,x=(-3-sqrt(265))/4
Solve for x: 1=log_{10}(2x^2+3x-22)
The solution is x=(-3+sqrt(265))/4 ,x=(-3-sqrt(265))/4