/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 100 Solve each equation. $$\ln 3-\... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve each equation. $$\ln 3-\ln (x+5)-\ln x=0$$

Short Answer

Expert verified
The solution to the equation is \( x = \frac{-5 + \sqrt{37}}{2} \).

Step by step solution

01

Combine Logarithms

Utilize the properties of logarithms to combine the logarithm terms. According to the properties, \( \ln a - \ln b = \ln \frac{a}{b}\). Therefore, the given equation \( \ln 3 - \ln (x+5) - \ln x = 0 \) becomes \( \ln 3 - \ln ((x+5)x) = 0 \) or \( \ln \frac{3}{(x+5)x} = 0\).
02

Convert Logarithmic Equation to Exponential Form

The property of logarithms states that \( \ln a = b \) is equivalent to \( e^b = a \). Therefore, convert the logarithmic equation \( \ln \frac{3}{(x+5)x} = 0 \) to the exponential form \( e^0 = \frac{3}{(x+5)x} \). We know that \( e^0 = 1 \), thus the equation simplifies to \( \frac{3}{(x+5)x} = 1 \).
03

Solve for x

Cross-multiply and set the equation equal to zero to solve for x. This gives us \( 3 = (x+5)x \), which simplifies to \( x^2 + 5x - 3 = 0 \). This is a quadratic equation, and we can solve for x using the quadratic formula \( x = {-b ± \sqrt{b^2 - 4ac}}/{2a} \). Substituting \( a = 1, b = 5, c = -3 \) gives us \( x = \frac{-5 ± \sqrt{5^2 - 4*1*-3}}{2*1} \), which simplifies to \( x = \frac{-5 ± \sqrt{25 + 12}}{2} \) or \( x = \frac{-5 ± \sqrt{37}}{2} \).
04

Check for extraneous solutions

Logarithms are only defined for positive numbers, so we need to check if the solutions \( x = \frac{-5 + \sqrt{37}}{2} \) and \( x = \frac{-5 - \sqrt{37}}{2} \) are valid. Since \( \frac{-5 - \sqrt{37}}{2} \) is negative, which makes \( \ln (x+5) \) undefined, it is not valid. The others solution is \( x = \frac{-5 + \sqrt{37}}{2} \), which is positive and within the range for which logarithms are defined.

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