What is the answer to log_{4}(x+1)-log_{4}(7-2x)=0 ?

The answer to log_{4}(x+1)-log_{4}(7-2x)=0 is x=2

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log_{4}(x+1)-log_{4}(7-2x)=0

lo g_{4} (x + 1) - lo g_{4} (7 - 2 x) = 0

x = 2

Solution steps

lo g_{4} (x + 1) - lo g_{4} (7 - 2 x) = 0

Add

lo g_{4} (7 - 2 x)

to both sides

lo g_{4} (x + 1) - lo g_{4} (7 - 2 x) + lo g_{4} (7 - 2 x) = 0 + lo g_{4} (7 - 2 x)

Simplify

lo g_{4} (x + 1) = lo g_{4} (7 - 2 x)

Apply log rules

x + 1 = 7 - 2 x

Solve

x + 1 = 7 - 2 x : x = 2

x = 2

Verify Solutions:

x = 2

True

The solution is

x = 2

lo g_{b} (x) + 2 = lo g_{b} (x + 2)

x, ln (y) = 9 ln (x + 15)

0 = ln (y)

d = 20 lo g_{10} (x)

ln (x) = 3.2614400000000 \dots

What is the answer to log_{4}(x+1)-log_{4}(7-2x)=0 ?
The answer to log_{4}(x+1)-log_{4}(7-2x)=0 is x=2
Solve for x: log_{4}(x+1)-log_{4}(7-2x)=0
The solution is x=2