What is the answer to log_{2}(x-2)=log_{2}(3+x)-1 ?

The answer to log_{2}(x-2)=log_{2}(3+x)-1 is x=7

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log_{2}(x-2)=log_{2}(3+x)-1

lo g_{2} (x - 2) = lo g_{2} (3 + x) - 1

x = 7

Solution steps

lo g_{2} (x - 2) = lo g_{2} (3 + x) - 1

Apply log rules

(x - 2) \cdot 2 = 3 + x

Solve

(x - 2) \cdot 2 = 3 + x : x = 7

x = 7

Verify Solutions:

x = 7

True

The solution is

x = 7

(lo g_{3} (x))^{2} = 4 lo g_{3} (x)

lo g_{a} (5) = - 2

lo g_{3} (x + 4) = lo g_{3} (1 - x)

lo g_{x} (\frac{1}{6}) = - 1

lo g_{7} (6 - 2 t) = 2

What is the answer to log_{2}(x-2)=log_{2}(3+x)-1 ?
The answer to log_{2}(x-2)=log_{2}(3+x)-1 is x=7
Solve for x: log_{2}(x-2)=log_{2}(3+x)-1
The solution is x=7