\section*{Problem 5 - 081045}

Determine all real numbers \(x\) that satisfy the equation \(4 \cdot \log _{4} x+3=2 \cdot \log _{x} 2\)!
}

With \(\log _{b} a=\frac{\log _{2} a}{\log _{2} b}\) and \(y:=\log _{2} x\), the equation transforms into \(\left[4 \cdot \frac{y}{2}+3=2 \cdot \frac{1}{y}\right.\), thus

\[
2 y^{2}+3 y-2=0 \quad \text { or } \quad y=-\frac{3}{4} \pm \sqrt{\frac{9}{16}+1}=\frac{3 \pm 5}{4}
\]

This yields the first solution \(y_{1}=\frac{-3+5}{4}=\frac{1}{2}\) and thus \(x_{1}=2^{y_{1}}=\sqrt{2}\) and the second \(y_{2}=\frac{-3-5}{4}=-2\) and hence \(x_{2}=2^{y_{2}}=\frac{1}{4}\).

Substituting into the original equation confirms both values.