What is the answer to log_{7}(2x^2-5x)=1 ?

The answer to log_{7}(2x^2-5x)=1 is x= 7/2 ,x=-1

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log_{7}(2x^2-5x)=1

lo g_{7} (2 x^{2} - 5 x) = 1

x = \frac{7}{2}, x = - 1

Decimal

x = 3.5, x = - 1

Solution steps

lo g_{7} (2 x^{2} - 5 x) = 1

Apply log rules

2 x^{2} - 5 x = 7

Solve

2 x^{2} - 5 x = 7 : x = \frac{7}{2}, x = - 1

x = \frac{7}{2}, x = - 1

lo g_{12} (x^{2} - 8 x + 16) = 0

lo g_{10} (x) 16 = 2

lo g_{x} (9) = 0.5

lo g_{10} (5 - 2 x) = 1

lo g_{8} (x) = \frac{2}{3}

What is the answer to log_{7}(2x^2-5x)=1 ?
The answer to log_{7}(2x^2-5x)=1 is x= 7/2 ,x=-1
Solve for x: log_{7}(2x^2-5x)=1
The solution is x= 7/2 ,x=-1