What is the answer to log_{2}(2x)-log_{2}(7x-15)=0 ?

The answer to log_{2}(2x)-log_{2}(7x-15)=0 is x=3

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log_{2}(2x)-log_{2}(7x-15)=0

lo g_{2} (2 x) - lo g_{2} (7 x - 15) = 0

x = 3

Solution steps

lo g_{2} (2 x) - lo g_{2} (7 x - 15) = 0

Add

lo g_{2} (7 x - 15)

to both sides

lo g_{2} (2 x) - lo g_{2} (7 x - 15) + lo g_{2} (7 x - 15) = 0 + lo g_{2} (7 x - 15)

Simplify

lo g_{2} (2 x) = lo g_{2} (7 x - 15)

Apply log rules

2 x = 7 x - 15

Solve

2 x = 7 x - 15 : x = 3

x = 3

Verify Solutions:

x = 3

True

The solution is

x = 3

lo g_{2} (7 x + 6) = 3

lo g_{4} (- x) + lo g_{4} (x + 10) = 2

lo g_{10} (x^{2} - 15 x) = 2

lo g_{2} (\frac{x}{6}) = 8

ln (3 x - 15) = 9

What is the answer to log_{2}(2x)-log_{2}(7x-15)=0 ?
The answer to log_{2}(2x)-log_{2}(7x-15)=0 is x=3
Solve for x: log_{2}(2x)-log_{2}(7x-15)=0
The solution is x=3