The number of real roots of the equation $$5 + 2^x -1 = 2^x \cdot 2^x - 2$$ is:

The given equation is

$$5 + 2^{x} \;=\; 2^{x}\cdot 2^{x} - 2.$$

First we simplify the right-hand side. We recall the law of indices

$$a^{m}\,a^{n}=a^{m+n}.$$

Using this with $$a=2,\; m=x,\; n=x,$$ we obtain

$$2^{x}\cdot 2^{x}=2^{x+x}=2^{2x}.$$

Substituting this back, the equation becomes

$$5 + 2^{x}=2^{2x}-2.$$

Now we collect all terms on one side:

$$2^{2x}-2^{x}-7=0.$$

At this point we make the substitution

$$t=2^{x}.$$

Because $$2^{x}>0$$ for every real $$x,$$ we have $$t>0.$$ With this substitution $$2^{2x}=(2^{x})^{2}=t^{2},$$ so the equation turns into

$$t^{2}-t-7=0.$$

This is a quadratic in $$t,$$ and we solve it with the quadratic-formula, which states

$$\text{For } at^{2}+bt+c=0,\quad t=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$$

Here $$a=1,\; b=-1,\; c=-7.$$ Substituting, we get

$$t=\dfrac{-(-1)\pm\sqrt{(-1)^{2}-4\cdot1\cdot(-7)}}{2\cdot1} =\dfrac{1\pm\sqrt{1+28}}{2} =\dfrac{1\pm\sqrt{29}}{2}.$$

Hence the two algebraic roots are

$$t_{1}=\dfrac{1+\sqrt{29}}{2},\qquad t_{2}=\dfrac{1-\sqrt{29}}{2}.$$

We must respect the condition $$t>0.$$ Clearly $$t_{1}>0$$ because both numerator and denominator are positive. However, $$\sqrt{29}\approx 5.385,$$ so

$$t_{2}=\dfrac{1-5.385}{2}\approx\dfrac{-4.385}{2}\approx-2.192<0.$$

This negative value cannot equal $$2^{x},$$ since $$2^{x}$$ is always positive. Therefore $$t_{2}$$ is rejected, and only

$$t=\dfrac{1+\sqrt{29}}{2}$$

remains.

We now return to the original variable. Because $$t=2^{x},$$ we have

$$2^{x}=\dfrac{1+\sqrt{29}}{2}\; \Longrightarrow\; x=\log_{2}\!\left(\dfrac{1+\sqrt{29}}{2}\right).$$

This single value of $$x$$ is real, and we have already shown that no other real value satisfies the equation. Hence there is exactly one real root.

Hence, the correct answer is Option C.