What is the answer to log_{4}(k^2-2k+1)=-1 ?

The answer to log_{4}(k^2-2k+1)=-1 is k= 3/2 ,k= 1/2

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log_{4}(k^2-2k+1)=-1

lo g_{4} (k^{2} - 2 k + 1) = - 1

k = \frac{3}{2}, k = \frac{1}{2}

Decimal

k = 1.5, k = 0.5

Solution steps

lo g_{4} (k^{2} - 2 k + 1) = - 1

Apply log rules

k^{2} - 2 k + 1 = \frac{1}{4}

Solve

k^{2} - 2 k + 1 = \frac{1}{4} : k = \frac{3}{2}, k = \frac{1}{2}

k = \frac{3}{2}, k = \frac{1}{2}

lo g_{x} (5) = - 1

lo g_{10} (x) = - 0.52

lo g_{10} (x^{2} + 9 x) = 1

\frac{ln ( x )}{ln ( x )} = 3

lo g_{3} (6 x - 2) = 2

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