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The solutions are:
1. \( x = 3 \)
2. \( x = \sqrt{5 + e} \)
Solución
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
\[
x=\sqrt{5+e}.
\]
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