\text{If a positive real}\mathrm{ x}\text{ satisfies }{\log }_{2}x+{\log }_{\sqrt{2}}x=48\text{, find the value of}x

The correct option is 2¹⁶.

We are given the logarithmic equation:

$${\log }_{2}x+{\log }_{\sqrt{2}}x=48$$

To solve this equation, we can use the change of base formula or properties of logarithms to express both terms with the same base. Recall the base-change property of logarithms:

$${\log }_{b^p}a=\frac{1}{p}{\log }_{b}a$$

In the second term of our equation, the base is $\sqrt{2}$, which can be written as $2^{\frac{1}{2}}$. Applying this property, we get:

$${\log }_{\sqrt{2}}x={\log }_{2^{\frac{1}{2}}}x=\frac{1}{\frac{1}{2}}{\log }_{2}x=2{\log }_{2}x$$

Now substitute this back into the original equation:

$${\log }_{2}x+2{\log }_{2}x=48$$

Combine the like terms on the left-hand side:

$$3{\log }_{2}x=48$$

Divide both sides of the equation by 3:

$${\log }_{2}x=16$$

To solve for x, rewrite the logarithmic equation in its exponential form:

$$x=2^{16}$$

Therefore, the value of x is 2¹⁶.