What is the answer to log_{4}(4x+5)-log_{4}(5-x)=1 ?

The answer to log_{4}(4x+5)-log_{4}(5-x)=1 is x= 15/8

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log_{4}(4x+5)-log_{4}(5-x)=1

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

x = \frac{15}{8}

Decimal Notation

x = 1.875

Solution steps

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

Add

lo g_{4} (5 - x)

to both sides

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

Simplify

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

Apply log rules

4 x + 5 = 4 (5 - x)

Solve

4 x + 5 = 4 (5 - x) : x = \frac{15}{8}

x = \frac{15}{8}

Verify Solutions:

x = \frac{15}{8}

True

The solution is

x = \frac{15}{8}

lo g_{2} (x) = - 0.9

lo g_{2} (x) = - 0.8

lo g_{2} (x) = - 0.5

lo g_{2} (x) = - 0.7

lo g_{10} (3) + lo g_{10} (x) = 2

What is the answer to log_{4}(4x+5)-log_{4}(5-x)=1 ?
The answer to log_{4}(4x+5)-log_{4}(5-x)=1 is x= 15/8
Solve for x: log_{4}(4x+5)-log_{4}(5-x)=1
The solution is x= 15/8