\( \left. \begin{array} { l l } { e ^ { 2 \ln ( 2 x - 3 ) } = 3 x } \\ { ( x ^ { 2 } - 5 ) ^ { \ln x } = x } & { 5.9 } \end{array} \right. \)

1. Solve the equation \[ e^{2\ln(2x-3)}=3x. \] Notice that using the property \[ e^{a\ln b}=(b)^a, \] we have \[ e^{2\ln(2x-3)}=(2x-3)^2. \] Thus the equation becomes \[ (2x-3)^2=3x. \] Expanding the left-hand side we get \[ 4x^2-12x+9=3x. \] Bringing all terms to one side yields \[ 4x^2-15x+9=0. \] The quadratic formula gives \[ x=\frac{15\pm\sqrt{15^2-4\cdot4\cdot9}}{2\cdot4}=\frac{15\pm\sqrt{225-144}}{8}=\frac{15\pm\sqrt{81}}{8}=\frac{15\pm9}{8}. \] Hence, \[ x=\frac{24}{8}=3 \quad \text{or} \quad x=\frac{6}{8}=\frac{3}{4}. \] However, since the original expression \(\ln(2x-3)\) is defined only when \[ 2x-3>0\quad\Longrightarrow\quad x>\frac{3}{2}, \] the only acceptable solution is \[ x=3. \] 2. Solve the equation \[ (x^2-5)^{\ln x}=x. \] First, for the expression \((x^2-5)^{\ln x}\) to be defined (using the usual real-exponentiation via logarithms), we require \[ x^2-5>0 \quad\Longrightarrow\quad x>\sqrt{5}\quad \text{or} \quad x<-\sqrt{5}. \] Since \(\ln x\) is defined only for \(x>0\), we must have \[ x>\sqrt{5}. \] Taking the natural logarithm on both sides of the equation (which is allowed as \(x>0\)) we get \[ \ln\Big((x^2-5)^{\ln x}\Big)=\ln x. \] Using the logarithm property \(\ln(a^b)=b\ln a\) (with the understanding that \(x^2-5>0\)), we obtain \[ (\ln x)\,\ln(x^2-5)=\ln x. \] There are two cases to consider: **Case 1: \(\ln x\neq 0\).** Divide both sides by \(\ln x\): \[ \ln(x^2-5)=1. \] Then \[ x^2-5=e, \] so \[ x^2=5+e. \] Since \(x>0\) we have \[ x=\sqrt{5+e}. \] We also note that \[ \sqrt{5+e}>\sqrt{5}, \] so this solution is acceptable. **Case 2: \(\ln x=0\).** This gives \[ x=1. \] However, for \(x=1\) the base becomes \[ x^2-5=1-5=-4, \] which is negative and the expression \((-4)^{\ln x}\) (with \(\ln x=0\)) is formally defined as \((-4)^0=1\), but the usual definition of \(a^b\) via \(e^{b\ln a}\) requires \(a>0\). Thus, under the standard definition and domain restrictions, \(x=1\) must be rejected. Therefore, the only solution for the second equation is \[ x=\sqrt{5+e}. \]