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

Question

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

UpStudy AI · Accepted Answer

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
   $$
   x=\sqrt{5+e}.
   $$
The solutions are: 1. $$ x = 3 $$ 2. $$ x = \sqrt{5 + e} $$

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

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