What is the answer to log_{4}(k)+log_{4}(k-3)=2 ?

The answer to log_{4}(k)+log_{4}(k-3)=2 is k=(3+sqrt(73))/2

Solve for k: log_{4}(k)+log_{4}(k-3)=2

The solution is k=(3+sqrt(73))/2

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lo g_{4} (k) + lo g_{4} (k - 3) = 2

k = \frac{3 + 73}{2}

Decimal Notation

k = 5.77200 \dots

Solution steps

lo g_{4} (k) + lo g_{4} (k - 3) = 2

Apply log rules

k (k - 3) = 16

Solve

k (k - 3) = 16 : k = \frac{3 + 73}{2}, k = \frac{3 - 73}{2}

k = \frac{3 + 73}{2}, k = \frac{3 - 73}{2}

Verify Solutions:

k = \frac{3 + 73}{2}

True

, k = \frac{3 - 73}{2}

False

The solution is

k = \frac{3 + 73}{2}

ln (x) = 1 + 3 ln (2)

90 = 10 lo g_{10} (x \cdot 1 0^{12})

lo g_{3} (x^{2} - 9) = 9

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

1 + 2 lo g_{5} (x) = lo g_{5} (16 x - 3)

What is the answer to log_{4}(k)+log_{4}(k-3)=2 ?
The answer to log_{4}(k)+log_{4}(k-3)=2 is k=(3+sqrt(73))/2
Solve for k: log_{4}(k)+log_{4}(k-3)=2
The solution is k=(3+sqrt(73))/2