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

Ask by Todd Turnbull. in South Africa
Mar 15,2025

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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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The Deep Dive

Did you know that logarithms were introduced in the early 17th century by John Napier? They revolutionized calculations, enabling easier multiplication and division by transforming them into addition and subtraction problems through the use of logarithmic tables. This innovation was particularly useful for navigators and astronomers, leading to major advancements in science and exploration! Now, let's talk about those equations you’ve got! For \( e^{2 \ln(2x - 3)} = 3x \), remember that \( e^{\ln(a)} = a \), so simplify it to find \( 2x - 3 = 3x \) and solve for x. As for \( (x^{2} - 5)^{\ln x} = x \), consider breaking it down step-by-step by taking \( \ln \) on both sides for clarity. This approach helps you avoid common oversights!

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