4 log x + 5 log x - 7 log x = log 16
Understand the Problem
The question involves solving a logarithmic equation that combines multiple logarithmic terms and sets them equal to another logarithm. The goal is to simplify and find the value of x that satisfies the equation.
Answer
The value of $x$ is $4$.
Answer for screen readers
The value of $x$ is $4$.
Steps to Solve
- Combine logarithmic terms
First, combine the logarithmic terms on the left side of the equation:
$$ 4 \log x + 5 \log x - 7 \log x = (4 + 5 - 7) \log x = 2 \log x $$
- Set up the simplified equation
Now the equation simplifies to:
$$ 2 \log x = \log 16 $$
- Use properties of logarithms
Apply the property of logarithms that states $a \log b = \log b^a$:
$$ \log x^2 = \log 16 $$
- Remove the logarithms
Since the logarithms are equal, we can remove them, resulting in:
$$ x^2 = 16 $$
- Solve for x
Taking the square root of both sides:
$$ x = \sqrt{16} = 4 $$
The value of $x$ is $4$.
More Information
In logarithmic equations, it’s crucial to remember that when two logarithms are equal, their arguments must also be equal. This property helps in simplifying and solving equations effectively.
Tips
- Not combining logarithmic terms correctly: Always ensure that coefficients are added or subtracted accurately.
- Failing to remove logarithms properly: Remember that if $\log a = \log b$, then $a$ must equal $b$.
AI-generated content may contain errors. Please verify critical information
