What is the answer to log_{2}(2w+5)-log_{2}(4-w)=3 ?

The answer to log_{2}(2w+5)-log_{2}(4-w)=3 is w= 27/10

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log_{2}(2w+5)-log_{2}(4-w)=3

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

w = \frac{27}{10}

Solution steps

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

Add

lo g_{2} (4 - w)

to both sides

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

Simplify

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

Apply log rules

2 w + 5 = 8 (4 - w)

Solve

2 w + 5 = 8 (4 - w) : w = \frac{27}{10}

w = \frac{27}{10}

Verify Solutions:

w = \frac{27}{10}

True

The solution is

w = \frac{27}{10}

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

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

x, ln (ln (x)) = 1

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

lo g_{4} (x - 6) = 1 - lo g_{4} (x - 3)

What is the answer to log_{2}(2w+5)-log_{2}(4-w)=3 ?
The answer to log_{2}(2w+5)-log_{2}(4-w)=3 is w= 27/10
Solve for w: log_{2}(2w+5)-log_{2}(4-w)=3
The solution is w= 27/10