What is the answer to log_{4}(4x)+log_{4}(2x)+8=4 ?

The answer to log_{4}(4x)+log_{4}(2x)+8=4 is x= 1/(32sqrt(2))

Solve for x: log_{4}(4x)+log_{4}(2x)+8=4

The solution is x= 1/(32sqrt(2))

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lo g_{4} (4 x) + lo g_{4} (2 x) + 8 = 4

x = \frac{1}{32 2}

Decimal Notation

x = 0.02209 \dots

Solution steps

lo g_{4} (4 x) + lo g_{4} (2 x) + 8 = 4

Apply log rules

x = \frac{1}{32 2}

Verify Solutions:

x = \frac{1}{32 2}

True

The solution is

x = \frac{1}{32 2}

ln (- x) = - ln (x)

lo g_{x} (16) = - 1

lo g_{x} (\frac{1}{36}) = - 2

lo g_{0.3} (x) = 2

lo g_{3} (x - 8) + lo g_{3} (x) = 2

What is the answer to log_{4}(4x)+log_{4}(2x)+8=4 ?
The answer to log_{4}(4x)+log_{4}(2x)+8=4 is x= 1/(32sqrt(2))
Solve for x: log_{4}(4x)+log_{4}(2x)+8=4
The solution is x= 1/(32sqrt(2))