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

The answer to log_{7}(x^2-x+1)=1 is x=3,x=-2

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

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

x = 3, x = - 2

Solution steps

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

Apply log rules

x^{2} - x + 1 = 7

Solve

x^{2} - x + 1 = 7 : x = 3, x = - 2

x = 3, x = - 2

0 = (ln (x))^{2}

lo g_{10} (x - 3) = 1 - lo g_{10} (x)

lo g_{10} (16 x) - lo g_{10} (2) = lo g_{10} (32)

lo g_{x} (\frac{1}{8}) = - 1.5

\frac{lo g _{10} ( 2 x )}{lo g _{10} ( 14 )} = 1

What is the answer to log_{7}(x^2-x+1)=1 ?
The answer to log_{7}(x^2-x+1)=1 is x=3,x=-2
Solve for x: log_{7}(x^2-x+1)=1
The solution is x=3,x=-2