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

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

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log_{2}(2ω+5)-log_{2}(4-ω)=3

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

ω = \frac{27}{10}

Solution steps

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

Add

lo g_{2} (4 - ω)

to both sides

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

Simplify

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

Apply log rules

2 ω + 5 = 8 (4 - ω)

Solve

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

ω = \frac{27}{10}

Verify Solutions:

ω = \frac{27}{10}

True

The solution is

ω = \frac{27}{10}

- ln (1 - x) = 46152

0.64 = lo g_{10} (y)

4 ln (4 x - 3) + 8 = 16

\frac{2 ( ln ( x ) ) + 2}{x} = 0

x, y = 9 ln (x - 4)

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