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

The answer to log_{4}(4x+2)-log_{4}(4-x)=1 is x= 7/4

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log_{4}(4x+2)-log_{4}(4-x)=1

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

x = \frac{7}{4}

Decimal

x = 1.75

Solution steps

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

Add

lo g_{4} (4 - x)

to both sides

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

Simplify

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

Apply log rules

4 x + 2 = 4 (4 - x)

Solve

4 x + 2 = 4 (4 - x) : x = \frac{7}{4}

x = \frac{7}{4}

Verify Solutions:

x = \frac{7}{4}

True

The solution is

x = \frac{7}{4}

lo g_{6} (a) = 3

lo g_{3} (x^{2} - 16) = 2

lo g_{4} (5 x) = 3

lo g_{5} (3 x) - 8 = 2

29.9 = 5 lo g_{10} (\frac{d}{10})

What is the answer to log_{4}(4x+2)-log_{4}(4-x)=1 ?
The answer to log_{4}(4x+2)-log_{4}(4-x)=1 is x= 7/4
Solve for x: log_{4}(4x+2)-log_{4}(4-x)=1
The solution is x= 7/4