Question

Solve for \(x\) a) \(2 \log _{3} x=\log _{3} 32+\log _{3} 2\) b) \(\frac{3}{2} \log _{7} x=\log _{7} 125\) c) \(\log _{2} x-\log _{2} 3=5\) d) \(\log _{6} x=2-\log _{6} 4\)

Short answer

a) x = 8b) x = 25c) x = 96d) x = 9

Step-by-step solution

Simplify the expression using logarithm properties

Given the equation: a) \(2 \log _{3} x=\log _{3} 32+\log _{3} 2\)Use the product rule of logarithms: \(\log_b a + \log_b c = \log_b (a \cdot c)\)So the equation becomes: \(2 \log _{3} x=\log _{3}(32 \cdot 2)\)This simplifies to: \(2 \log _{3} x=\log _{3} 64\)

Isolate the variable

Divide both sides by 2:\(\log _{3} x = \frac{\log _{3} 64}{2}\)This simplifies to: \(\log _{3} x = \log _{3} 8\)

Solve for x

Since the bases and the logs are the same:\(x = 8\)

Convert logarithmic expressions using properties

Given the equation: b) \(\frac{3}{2} \log _{7} x = \log _{7} 125\)

Isolate the variable

Multiply both sides by \(\frac{2}{3}\):\(\log _{7} x = \frac{2}{3} \log _{7} 125\)

Use exponent rules

We know that \(\frac{2}{3} \log _{7} 125\) is the same as \(\log _{7} 125^{\frac{2}{3}}\):\(\log _{7} x = \log _{7} 25\)

Solve for x

Since the bases and the logs are the same:\(x = 25\)

Convert logarithmic expressions using properties

Given the equation: c) \(\log _{2} x - \log _{2} 3 = 5\)

Use quotient rule

Combine using the quotient rule: \(\log _{2} \left( \frac{x}{3} \right) = 5\)

Remove the logarithm

Rewrite the equation in exponential form: \(\frac{x}{3} = 2^{5}\)

Solve for x

Multiply both sides by 3: \(x = 3 \cdot 32\)So: \(x = 96\)

Convert right-hand side using properties

Given the equation: d) \(\log _{6} x = 2 - \log _{6} 4\)

Use subtraction property

Combine using the subtraction property: \(\log _{6} x = \log _{6} 6^{2} - \log _{6} 4\)Which simplifies to: \(\log _{6} x = \log _{6} \left( \frac{36}{4} \right)\)

Simplify the argument

Simplify \(\frac{36}{4}\) to: \(\log _{6} x = \log _{6} 9\)

Solve for x

Since the bases and the logs are the same:\(x = 9\)

Key concepts

logarithm properties

Logarithms help us to solve for unknown exponents and manage multiplication and division within logarithmic expressions. Some essential properties include:
- **Product Rule**: \(\text{log}_b(a \times c) = \text{log}_b(a) + \text{log}_b(c)\). This enables us to compress the sum of two logs into a single logarithm.
- **Quotient Rule**: \(\text{log}_b \left( \frac{a}{c} \right) = \text{log}_b(a) - \text{log}_b(c)\). This is handy for combining differences into a single logarithm.
- **Power Rule**: \(\text{log}_b(a^c) = c \times \text{log}_b(a)\). This brings the exponent out as a coefficient.
Understanding these properties is fundamental. For instance, \(2 \text{log}_3(x) = \text{log}_3(64)\) can be solved with these rules. We rewrite \(2 \text{log}_3(x)\) as \(\text{log}_3(x^2)\), simplifying to \(\text{log}_3(64)\). Then, \(x^2 = 64\) leads us to \(x = 8\).

solving logarithm equations

Solving logarithm equations involves using logarithm properties to isolate the variable. Here is the general process:
1. **Apply Logarithm Properties**: Simplify the given logarithmic expressions using product, quotient, or power rules.
2. **Isolate the Logarithm**: Rearrange the equation to isolate the logarithmic expression on one side.
3. **Convert to Exponential Form**: If needed, rewrite the logarithmic equation in its exponential form to solve for the variable.
Take the equation \(\frac{3}{2} \text{log}_7(x) = \text{log}_7(125)\). We isolate \(\text{log}_7(x)\) by multiplying both sides by \(\frac{2}{3}\). This transforms the equation into \(\text{log}_7(x) = \text{log}_7(25)\). Since the log bases and values are equal, it follows that \(x = 25\).

logarithm rules

Logarithm rules allow us to manipulate logarithmic expressions effectively. They include:
- **Change of Base Rule**: Helps converting logarithms of different bases, useful if calculators have specific log functions.
- **Single Logarithm Rule**: Helps compress multiple logarithms into one, which simplifies equations.
Consider \(\text{log}_2(x) - \text{log}_2(3) = 5\). Using the quotient rule, rewrite it to get \(\text{log}_2\left( \frac{x}{3} \right) = 5\). Then, convert to exponential form \(\frac{x}{3} = 2^5\), which becomes \(\frac{x}{3} = 32\). Solve for \(x\) by multiplying both sides by 3, resulting in \(x = 96\).

exponential form

Rewriting logarithms in exponential form is crucial. It helps in solving complex logarithmic equations. The general transform is \(\text{log}_b(a) = c\) converts to \(b^c = a\).
For example, with \(\text{log}_6(x) = 2 - \text{log}_6(4)\), use subtraction rule to combine terms: \(\text{log}_6(x) = \text{log}_6 \left( \frac{36}{4} \right)\), simplifying to \(\text{log}_6(x) = \text{log}_6(9)\). Since the logs are equal, \(x = 9\). The conversion to exponential form helps to isolate and solve for the variable directly when dealing with more complex logarithmic components.