Solve the following equation

$$
3^{2 x}+3^{2 x-1}=2^{x+\frac{7}{2}}+2^{x+\frac{1}{2}}
$$
I. Solution: We factor out the second term on both sides of the equation:

$$
3^{2 x-1}(3+1)=2^{x+\frac{1}{2}}\left(2^{3}+1\right)
$$

which is

$$
\left(3^{2}\right)^{x-\frac{1}{2}} \cdot 2^{2}=2^{x-\frac{1}{3}} \cdot 3^{2}
$$

Dividing both sides by $3^{2} \cdot 2^{2}$

$$
9^{x-\frac{3}{2}}=2^{x-\frac{3}{2}}
$$

The powers of two different bases can only be equal if the exponent is 0, hence

$$
x=\frac{3}{2}
$$

László Bognár (Veszprém, Lovassy g. II. o. t.)

II. Solution: By factoring out from the equation, we get

$$
3^{2 x}\left(1+\frac{1}{2}\right)=2^{x} \cdot 2^{\frac{1}{3}} \cdot\left(2^{3}+1\right)
$$

which is

$$
\left(\frac{3^{2}}{2}\right)^{x}=\left(\frac{9}{2}\right)^{x}=\frac{3^{3}}{2^{\frac{3}{2}}}=\frac{\left(3^{2}\right)^{\frac{3}{2}}}{2^{\frac{3}{2}}}=\left(\frac{9}{2}\right)^{\frac{3}{2}}
$$

From this, it is clear that

$$
x=\frac{3}{2}
$$

Mária Csernyánszky (Bp. XVI., Corvin Mátyás g. I. o.)