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

The answer to log_{4}(x+2)+log_{2}(4)=0 is x=-31/16

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log_{4}(x+2)+log_{2}(4)=0

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

x = - \frac{31}{16}

Decimal Notation

x = - 1.9375

Solution steps

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

Move

lo g_{2} (4)

to the right side

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

Apply log rules

x + 2 = \frac{1}{16}

Solve

x + 2 = \frac{1}{16} : x = - \frac{31}{16}

x = - \frac{31}{16}

Verify Solutions:

x = - \frac{31}{16}

True

The solution is

x = - \frac{31}{16}

lo g_{3} (x - 4) + lo g_{3} (x + 6) = 1

lo g_{x} (1.41) = 0.5

- 3 = 20 lo g_{10} (x)

lo g_{5} (x + 1) = 2 + lo g_{5} (x - 1)

e^{2} ln (b) = 400

What is the answer to log_{4}(x+2)+log_{2}(4)=0 ?
The answer to log_{4}(x+2)+log_{2}(4)=0 is x=-31/16
Solve for x: log_{4}(x+2)+log_{2}(4)=0
The solution is x=-31/16