/*! 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 121 Solve the equation. \(\log w+4... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 121

Solve the equation. \(\log w+4 \sqrt{\log w}-12=0\)

Short Answer

Expert verified
The solution is \( w = 10,000 \).

Step by step solution

01

Substitute a variable

Let \( x = \sqrt{\log w} \). Consequently, the original equation \( \log w + 4 \sqrt{\log w} - 12 = 0 \) can be rewritten using this substitution. Since \( \log w = x^2 \), the equation becomes \( x^2 + 4x - 12 = 0 \).
02

Solve the quadratic equation

The equation \( x^2 + 4x - 12 = 0 \) is a standard quadratic equation in the form \( ax^2 + bx + c = 0 \). Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( x \). Here, \( a = 1 \), \( b = 4 \), and \( c = -12 \). Thus:\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 + 48}}{2} = \frac{-4 \pm \sqrt{64}}{2} = \frac{-4 \pm 8}{2} \]
03

Simplify the solutions

Simplify the two possible solutions from the quadratic formula:\[ x = \frac{-4 + 8}{2} = 2 \]\[ x = \frac{-4 - 8}{2} = -6 \]Since \( x = \sqrt{\log w} \) represents a square root, \( x \) must be non-negative. Therefore, \( x = -6 \) is not valid and only \( x = 2 \) is a solution.
04

Solve for \( w \)

Recall the substitution \( x = \sqrt{\log w} \). Since \( x = 2 \), then \( 2 = \sqrt{\log w} \). Squaring both sides, the equation becomes \( 4 = \log w \). To determine \( w \), rewrite the logarithmic equation in exponential form:\[ 10^4 = w \]Thus, \( w = 10,000 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

quadratic equations in algebra
Quadratic equations are a fundamental concept in algebra. They are equations of the form \[ ax^2 + bx + c = 0 \], where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. In this exercise, the equation \( x^2 + 4x - 12 = 0 \) is a quadratic equation. To solve for \( x \), we use methods like factorization, completing the square, or the quadratic formula. In our case, we used the quadratic formula because it is straightforward and works for any quadratic equation. Remember that solving quadratic equations allows us to find the values of \( x \) that satisfy the equation.
logarithmic substitution
Logarithmic substitution is a technique used to simplify equations involving logarithms. By substituting part of the equation with a new variable, we can transform a complex logarithmic equation into a more manageable form. In the given exercise, we let \( x = \sqrt{\log w} \). This substitution simplifies the equation \( \log w + 4 \sqrt{\log w} - 12 = 0 \) to \( x^2 + 4x - 12 = 0 \). This is easier to solve because it transforms the problem from dealing with a logarithmic equation to solving a quadratic equation.
quadratic formula application
The quadratic formula is an essential tool for solving quadratic equations. It states that the solutions for \( ax^2 + bx + c = 0 \) can be found using \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. In our problem, we identified \( a = 1 \), \( b = 4 \), and \( c = -12 \). Plugging these values into the formula, we get: \[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \]. Simplifying under the square root, we find \[ x = \frac{-4 \pm \sqrt{64}}{2} = \frac{-4 \pm 8}{2} \]. This gives two potential solutions: \( x = 2 \) and \( x = -6 \). However, because \( x \) involves a square root, only the non-negative solution \( x = 2 \) is valid.
exponential and logarithmic relationships
Understanding the relationship between exponential and logarithmic functions is crucial. The logarithm function \( \log_b(y) = x \) is the inverse of the exponential function \( b^x = y \). This means that solving \( \log_b(y) = x \) for \( y \) is equivalent to finding \( y = b^x \). In our exercise, we needed to solve for \( w \) after determining that \( \sqrt{\log w} = 2 \). By squaring both sides, we found \( \log w = 4 \). Rewriting this in exponential form, we get \( w = 10^4 \), leading to \( w = 10,000 \). This demonstrates the power of understanding how logarithms and exponentials are interconnected.

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