The number of solutions of the equation $$32^{\tan^2 x} + 32^{\sec^2 x} = 81$$, $$0 \leq x \leq \frac{\pi}{4}$$ is:

We start with the given transcendental equation

$$32^{\tan^2x}+32^{\sec^2x}=81,\qquad 0\le x\le\frac{\pi}{4}.$$

First we recall the Pythagorean identity for tangent and secant:

$$\sec^2x\,=\,1+\tan^2x.$$

Let us put

$$t=\tan^2x.$$

Because $$0\le x\le\frac{\pi}{4}$$, we have $$0\le \tan x\le1$$; hence

$$0\le t\le1.$$

Using $$\sec^2x=1+t$$ in the original equation gives

$$32^{t}+32^{\,1+t}=81.$$

By the property of exponents $$a^{m+n}=a^{m}\,a^{n}$$, we write

$$32^{\,1+t}=32^{1}\cdot32^{t}=32\;32^{t}.$$

Substituting this result back, the equation becomes

$$32^{t}+32\;32^{t}=81.$$

Taking $$32^{t}$$ common we obtain

$$32^{t}\bigl(1+32\bigr)=81.$$

Since $$1+32=33$$, we have

$$32^{t}\,.\,33=81.$$

Dividing both sides by $$33$$:

$$32^{t}=\frac{81}{33}=\frac{27}{11}.$$

Now $$32=2^{5}$$. Hence

$$32^{t}=(2^{5})^{t}=2^{5t}=\frac{27}{11}.$$

Taking base-$$2$$ logarithms on both sides (formula: if $$a^b=c$$, then $$b=\log_{a}c$$):

$$5t=\log_{2}\!\Bigl(\tfrac{27}{11}\Bigr).$$

Therefore

$$t=\frac{1}{5}\,\log_{2}\!\Bigl(\tfrac{27}{11}\Bigr).$$

Because $$\tfrac{27}{11}\approx2.4545$$ and $$\log_{2}(2.4545)\approx1.296$$, we get a numerical value

$$t\approx\frac{1.296}{5}\approx0.259.$$

Remember $$t=\tan^{2}x$$. So

$$\tan^{2}x\approx0.259.$$

Taking the positive square root (only the positive root is relevant here since $$\tan x\ge0$$ on the interval $$0\le x\le\frac{\pi}{4}$$):

$$\tan x\approx\sqrt{0.259}\approx0.509.$$

Because the tangent function is strictly increasing on $$\bigl[0,\tfrac{\pi}{4}\bigr]$$, the equation $$\tan x\approx0.509$$ possesses exactly one solution in that interval. Consequently the original equation has exactly one admissible $$x$$ between $$0$$ and $$\tfrac{\pi}{4}$$.

Hence, the correct answer is Option C.