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Magazine Chapter 4: Problem 40
Determine if the given value of \(x\) is a solution to the logarithmic equation. \(\log _{4} x=3-\log _{4}(x-63)\) a. \(x=64\) b. \(x=-1\) c. \(x=32\)
Short Answer
Expert verified
a. x=64
Step by step solution
01
Rewrite the equation in terms of a single logarithm
Start by combining the logarithmic terms on one side. Use the logarithmic property: \text{If \(\log_a b - \log_a c = \log_a \left(\frac{b}{c}\right)\)}\text , then the given equation can be rewritten as: \log_4 x + \log_4 (x - 63) = 3.
02
Combine the logarithms
Combine the two logarithms using the product property: \text{If \(\log_a b + \log_a c = \log_a (bc)\), then the equation becomes: \log_4 (x(x - 63)) = 3.}
03
Exponentiate both sides
Rewrite the equation in exponential form: \(x(x - 63) = 4^3\). Simplify the right side: 4^3=64. So the equation is: x^2 - 63x = 64.
04
Bring the equation to standard quadratic form
Rewrite the equation as: x^2 - 63x - 64 = 0.
05
Solve the quadratic equation
Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 1\), \(b = -63\), and \(c = -64\). Calculate the discriminant: \(b^2 - 4ac = 63^2 + 4*64 = 3969 + 256 = 4225\). So, the roots are \(x = \frac{63 \pm \sqrt{4225}}{2}\). Since \(\sqrt{4225} = 65\), we get the roots: \(x = \frac{63 + 65}{2} = 64\) and \(x = \frac{63 - 65}{2} = -1\).
06
Check the validity of the roots
Since the original logarithmic expressions must be defined, \(x\) and \(x-63\) must be positive. Check both solutions to see which are valid: \(x = 64\) is valid since both 64 and 1 (64 - 63) are positive. \(x = -1\) is not valid since both -1 and -64 are not positive.
07
Final answer
The value \(x = 64\) is the solution, while \(x = -1\) and \(x = 32\) are not solutions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
logarithmic properties
Understanding logarithmic properties is crucial for solving logarithmic equations. One essential property is *logarithm subtraction*. For example, \( \log_a b - \log_a c = \log_a \left( \frac{b}{c} \right) \). This property allows us to combine or simplify logarithmic expressions, often making the equation easier to solve.
Another key property is the *logarithm product*. It states \(\log_a b + \log_a c = \log_a (bc)\). This property helps when we need to combine logarithms into a single term or expand them to separate terms.
Using these properties, we can manipulate logarithmic equations to find variables more effectively.
Another key property is the *logarithm product*. It states \(\log_a b + \log_a c = \log_a (bc)\). This property helps when we need to combine logarithms into a single term or expand them to separate terms.
Using these properties, we can manipulate logarithmic equations to find variables more effectively.
quadratic equation
A quadratic equation has the general form \(ax^2 + bx + c = 0\). To solve it, we often use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Let's break this down:
- **Identifying coefficients**: The coefficients are *a*, *b*, and *c* from the equation.
- **Calculating the discriminant**: The discriminant is \(b^2 - 4ac \). It helps to determine the nature of the roots. If the discriminant is positive, we get two real roots. If zero, one real root, and if negative, no real roots.
- **Applying the quadratic formula**: By substituting *a*, *b*, and *c* into the formula, we find the roots (solutions) of the quadratic equation.
For example, in our solution, the quadratic equation derived was \(x^2 - 63x - 64 = 0\). Plugging in the values, we found two solutions: \(x = 64\) and \(x = -1\), though only \(x = 64\) was valid after checking.
- **Identifying coefficients**: The coefficients are *a*, *b*, and *c* from the equation.
- **Calculating the discriminant**: The discriminant is \(b^2 - 4ac \). It helps to determine the nature of the roots. If the discriminant is positive, we get two real roots. If zero, one real root, and if negative, no real roots.
- **Applying the quadratic formula**: By substituting *a*, *b*, and *c* into the formula, we find the roots (solutions) of the quadratic equation.
For example, in our solution, the quadratic equation derived was \(x^2 - 63x - 64 = 0\). Plugging in the values, we found two solutions: \(x = 64\) and \(x = -1\), though only \(x = 64\) was valid after checking.
exponential equations
Exponential equations involve expressions with variables in the exponent. To solve them, follow these steps:
- **Rewrite in logarithmic form**: If you have an equation like \(4^3 = 64\), recognize that 64 can be written as \(4^3\). This simplifies the comparison.
- **Use logarithms**: Apply logarithms to both sides if the exponents are not straightforward. This helps in pulling down the exponent.
- **Equate the exponents**: Once the bases match, equate the exponents and solve for the variable.
In our given problem, we rewrote \(\log_4 (x(x - 63)) = 3\) in exponential form as \(x(x - 63) = 4^3\), simplifying further to solve the quadratic equation.
- **Rewrite in logarithmic form**: If you have an equation like \(4^3 = 64\), recognize that 64 can be written as \(4^3\). This simplifies the comparison.
- **Use logarithms**: Apply logarithms to both sides if the exponents are not straightforward. This helps in pulling down the exponent.
- **Equate the exponents**: Once the bases match, equate the exponents and solve for the variable.
In our given problem, we rewrote \(\log_4 (x(x - 63)) = 3\) in exponential form as \(x(x - 63) = 4^3\), simplifying further to solve the quadratic equation.
