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

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

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

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

x = 1

Solution steps

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

Apply log rules

(x + 1)^{2} = 4 x

Solve

(x + 1)^{2} = 4 x : x = 1

x = 1

Verify Solutions:

x = 1

True

The solution is

x = 1

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

ln (10 x + 5) - ln (4 - x) = ln (2)

lo g_{10} (x + 5) - 9 = lo g_{10} (3 x)

lo g_{x} (64) = \frac{2}{3}

ln (4 x) - ln (x + 3) = ln (x^{2})

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