Question

Solve for \(x\) a) \(\log _{4} x+\log _{2} x=6\) b) \(\log _{3} x-\log _{27} x=\frac{4}{3}\)

Short answer

a) \( x = 16 \). b) \( x = 9 \).

Step-by-step solution

Title - Rewrite using log properties (Part a)

Rewrite the given equation \( \log_{4} x + \log_{2} x = 6 \) using properties of logarithms. We know \( \log_b a + \log_b c = \log_b (ac) \) and \( \log_{4} x = \frac{\log_{2} x}{\log_{2} 4} = \frac{\log_{2} x}{2} \). This gives us \( \frac{\log_{2} x}{2} + \log_{2} x = 6 \).

Title - Combine and simplify logs (Part a)

Combine the logarithmic terms: \[ \frac{\log_{2} x}{2} + \log_{2} x = \frac{\log_{2} x}{2} + \frac{2 \log_{2} x}{2} = \frac{3 \log_{2} x}{2} \]. Thus, \[ \frac{3 \log_{2} x}{2} = 6 \].

Title - Solve for \( \log_{2} x \) (Part a)

Solve the equation: \[ \frac{3 \log_{2} x}{2} = 6 \Rightarrow 3 \log_{2} x = 12 \Rightarrow \log_{2} x = 4 \].

Title - Solve for \( x \) (Part a)

Convert the logarithmic equation to exponent form: \( \log_{2} x = 4 \Rightarrow x = 2^4 = 16 \). The solution for the first part is \( x = 16 \).

Title - Rewrite using log properties (Part b)

Rewrite the given equation \( \log_{3} x - \log_{27} x = \frac{4}{3} \). Using the change of base formula, \( \log_{27} x = \frac{\log_{3} x}{\log_{3} 27} = \frac{\log_{3} x}{3} \) (since \( \log_{3} 27 = 3 \)). This yields \( \log_{3} x - \frac{\log_{3} x}{3} = \frac{4}{3} \).

Title - Combine and simplify logs (Part b)

Combine the logarithmic terms: \[ \log_{3} x - \frac{\log_{3} x}{3} = \frac{3 \log_{3} x - \log_{3} x}{3} = \frac{2 \log_{3} x}{3} \]. Thus, \[ \frac{2 \log_{3} x}{3} = \frac{4}{3} \].

Title - Solve for \( \log_{3} x \) (Part b)

Solve the equation: \[ \frac{2 \log_{3} x}{3} = \frac{4}{3} \Rightarrow 2 \log_{3} x = 4 \Rightarrow \log_{3} x = 2 \].

Title - Solve for \( x \) (Part b)

Convert the logarithmic equation to exponent form: \( \log_{3} x = 2 \Rightarrow x = 3^2 = 9 \). The solution for the second part is \( x = 9 \).

Key concepts

Properties of Logarithms

Understanding the properties of logarithms can be immensely helpful when solving logarithmic equations. One key property is the *product rule*, which states that \(\begin{array}{c}\text{log}_b(a) + \text{log}_b(c) = \text{log}_b(ac) \end{array}\). This allows you to combine multiple logarithmic terms into one. Another important property is the *quotient rule*: \(\begin{array}{c}\text{log}_b(a) - \text{log}_b(c) = \text{log}_b\frac{a}{c} \end{array}\). Additionally, there's the *power rule*, which states that \(\begin{array}{c}\text{log}_b(a^c) = c\text{log}_b(a) \end{array}\). These properties are foundational tools for manipulating logarithmic expressions. For example, using these properties, you can simplify complex logarithmic equations and make them more manageable.

Logarithmic Identities

Logarithms have certain identities which are very useful when solving equations. One such identity is \(\begin{array}{c}\text{log}_b(b) = 1 \end{array}\), which helps in simplifying expressions when the base and the argument are the same. Another identity is \(\begin{array}{c}\text{log}_b(1) = 0 \end{array}\), useful for zeroing out terms. These identities often assist in reducing the calculation complexity and can instantly simplify parts of equations.

Exponentiation

Exponentiation is the inverse operation of taking logarithms. For example, if \(\begin{array}{c}\text{log}_b(x) = y \end{array}\) then \(\begin{array}{c}b^y = x \end{array}\). In the context of solving logarithmic equations, once you have isolated the logarithmic term, you can convert it back to an exponential form to solve for the unknown variable. This method is particularly helpful when you have a result such as \(\begin{array}{c}\text{log}_b(x) = y \end{array}\) and need to find \(\begin{array}{c}x \end{array}\).

Change of Base Formula

The change of base formula is an essential tool for dealing with logarithms that have bases other than 10 or *e*. The formula is expressed as \(\begin{array}{c}\text{log}_b(a) = \frac{\text{log}_c(a)}{\text{log}_c(b)} \end{array}\), where \(\begin{array}{c}c\) is a new base. This is particularly helpful when you need to compare or combine logarithms with different bases. For instance, when solving problems involving different bases, converting everything to a common base can simplify the equation significantly.