What is the answer to log_{3}(4x+4)-log_{3}(2-x)=1 ?

The answer to log_{3}(4x+4)-log_{3}(2-x)=1 is x= 2/7

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log_{3}(4x+4)-log_{3}(2-x)=1

lo g_{3} (4 x + 4) - lo g_{3} (2 - x) = 1

x = \frac{2}{7}

Solution steps

lo g_{3} (4 x + 4) - lo g_{3} (2 - x) = 1

Add

lo g_{3} (2 - x)

to both sides

lo g_{3} (4 x + 4) - lo g_{3} (2 - x) + lo g_{3} (2 - x) = 1 + lo g_{3} (2 - x)

Simplify

lo g_{3} (4 x + 4) = 1 + lo g_{3} (2 - x)

Apply log rules

4 x + 4 = 3 (2 - x)

Solve

4 x + 4 = 3 (2 - x) : x = \frac{2}{7}

x = \frac{2}{7}

Verify Solutions:

x = \frac{2}{7}

True

The solution is

x = \frac{2}{7}

ln (x + 2) - ln (x - 1) = ln (4)

lo g_{8} (16) + lo g_{8} (y - 2) = 2

lo g_{a} (x) = b

4 lo g_{3} (x) - 11 = 5

0 = ln (c)

What is the answer to log_{3}(4x+4)-log_{3}(2-x)=1 ?
The answer to log_{3}(4x+4)-log_{3}(2-x)=1 is x= 2/7
Solve for x: log_{3}(4x+4)-log_{3}(2-x)=1
The solution is x= 2/7