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

The answer to 2log_{2}(x)+log_{2}(a)=0 is x=2^{-(log_{2}(a))/2}

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

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

x = 2^{- \frac{l o g _{2} ( a )}{2}}

Solution steps

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

Move

lo g_{2} (a)

to the right side

2 lo g_{2} (x) = - lo g_{2} (a)

Divide both sides by

2

lo g_{2} (x) = - \frac{lo g _{2} ( a )}{2}

Apply log rules

x = 2^{- \frac{l o g _{2} ( a )}{2}}

Verify Solutions:

x = 2^{- \frac{l o g _{2} ( a )}{2}}

True

The solution is

x = 2^{- \frac{l o g _{2} ( a )}{2}}

6 = - lo g_{10} (x)

lo g_{3} (p) = 6

0.33 = lo g_{10} (x)

4 lo g_{4} (x) = lo g_{x} (4)

x, ln (y) = 4 ln (x)

What is the answer to 2log_{2}(x)+log_{2}(a)=0 ?
The answer to 2log_{2}(x)+log_{2}(a)=0 is x=2^{-(log_{2}(a))/2}