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

The answer to log_{2}(x)+log_{4}(x)=1 is x=\sqrt[3]{4}

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

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

Decimal Notation

x = 1.58740 \dots

Solution steps

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

Apply log rules

Verify Solutions:

True

The solution is

lo g_{2 a} (4) = 2

lo g_{10} (x) + lo g_{10} (x + 3) = 2

lo g_{10} (x) \frac{1}{25} = - 2

0.367 = - lo g_{10} (x)

x, lo g_{x} (64) = - 2

What is the answer to log_{2}(x)+log_{4}(x)=1 ?
The answer to log_{2}(x)+log_{4}(x)=1 is x=\sqrt[3]{4}
Solve for x: log_{2}(x)+log_{4}(x)=1
The solution is x=\sqrt[3]{4}