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

The answer to log_{2}(4x+5)-log_{2}(4-x)=3 is x= 9/4

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

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

x = \frac{9}{4}

Solution steps

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

Add

lo g_{2} (4 - x)

to both sides

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

Simplify

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

Apply log rules

4 x + 5 = 8 (4 - x)

Solve

4 x + 5 = 8 (4 - x) : x = \frac{9}{4}

x = \frac{9}{4}

Verify Solutions:

x = \frac{9}{4}

True

The solution is

x = \frac{9}{4}

ln (x) = ln^{2} (x)

lo g_{6} (5 x) = 3

lo g_{10} (x) + lo g_{10} (x + 90) = 3

t, ln (2 t + 3) = 4

1 = 2.22 + lo g_{10} (x)

What is the answer to log_{2}(4x+5)-log_{2}(4-x)=3 ?
The answer to log_{2}(4x+5)-log_{2}(4-x)=3 is x= 9/4
Solve for x: log_{2}(4x+5)-log_{2}(4-x)=3
The solution is x= 9/4