What is the answer to 2log_{10}(x)+log_{10}(4)=-2 ?

The answer to 2log_{10}(x)+log_{10}(4)=-2 is x=10^{(-2-2log_{10}(2))/2}

Solve for x: 2log_{10}(x)+log_{10}(4)=-2

The solution is x=10^{(-2-2log_{10}(2))/2}

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2 lo g_{10} (x) + lo g_{10} (4) = - 2

x = 1 0^{\frac{- 2 - 2 l o g _{10} ( 2 )}{2}}

Decimal

x = 0.05

Solution steps

2 lo g_{10} (x) + lo g_{10} (4) = - 2

Move

lo g_{10} (4)

to the right side

2 lo g_{10} (x) = - 2 - lo g_{10} (4)

Divide both sides by

2

lo g_{10} (x) = \frac{- 2 - lo g _{10} ( 4 )}{2}

Apply log rules

x = 1 0^{\frac{- 2 - 2 l o g _{10} ( 2 )}{2}}

Verify Solutions:

x = 1 0^{\frac{- 2 - 2 l o g _{10} ( 2 )}{2}}

True

The solution is

x = 1 0^{\frac{- 2 - 2 l o g _{10} ( 2 )}{2}}

lo g_{x + 5} (4) = 2

- lo g_{10} (x) = 2.80524

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

lo g_{x} (2500) = 2

lo g_{4} (5 x + 9) = 3

What is the answer to 2log_{10}(x)+log_{10}(4)=-2 ?
The answer to 2log_{10}(x)+log_{10}(4)=-2 is x=10^{(-2-2log_{10}(2))/2}
Solve for x: 2log_{10}(x)+log_{10}(4)=-2
The solution is x=10^{(-2-2log_{10}(2))/2}