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

The answer to log_{2}(y+2)-log_{2}(y-2)=1 is y=6

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log_{2}(y+2)-log_{2}(y-2)=1

lo g_{2} (y + 2) - lo g_{2} (y - 2) = 1

y = 6

Solution steps

lo g_{2} (y + 2) - lo g_{2} (y - 2) = 1

Add

lo g_{2} (y - 2)

to both sides

lo g_{2} (y + 2) - lo g_{2} (y - 2) + lo g_{2} (y - 2) = 1 + lo g_{2} (y - 2)

Simplify

lo g_{2} (y + 2) = 1 + lo g_{2} (y - 2)

Apply log rules

y + 2 = 2 (y - 2)

Solve

y + 2 = 2 (y - 2) : y = 6

y = 6

Verify Solutions:

y = 6

True

The solution is

y = 6

4 ln (x + 6) = - 8

lo g_{3} (x + a) = b

lo g_{3} (4 x - 5) - lo g_{3} (x + 2) = 1

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

- 33.10906 = 10 \cdot lo g_{10} (x)

What is the answer to log_{2}(y+2)-log_{2}(y-2)=1 ?
The answer to log_{2}(y+2)-log_{2}(y-2)=1 is y=6
Solve for y: log_{2}(y+2)-log_{2}(y-2)=1
The solution is y=6