Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Get quick and reliable answers to your questions from a dedicated community of professionals on our platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
To solve the logarithmic equation
[tex]\[ \log_2(x + 2) = 1 + \log_2(x - 5), \][/tex]
follow these steps:
### Step 1: Understand the Equation
First, recognize that the logarithms are both to the base 2.
[tex]\[ \log_2(x + 2) \][/tex]
[tex]\[ \log_2(x - 5) + 1 \][/tex]
### Step 2: Isolate the Logarithmic Terms
To simplify the equation, subtract \(\log_2(x - 5)\) from both sides:
[tex]\[ \log_2(x + 2) - \log_2(x - 5) = 1 \][/tex]
### Step 3: Apply Logarithmic Properties
Use the property of logarithms that states \(\log_b(a) - \log_b(c) = \log_b\left( \frac{a}{c} \right)\):
[tex]\[ \log_2 \left(\frac{x + 2}{x - 5}\right) = 1 \][/tex]
### Step 4: Eliminate the Logarithm
Remember, if \(\log_b(A) = C\), then \(A = b^C\). In this case, \(b = 2\) and \(C = 1\):
[tex]\[ \frac{x + 2}{x - 5} = 2^1 \][/tex]
[tex]\[ \frac{x + 2}{x - 5} = 2 \][/tex]
### Step 5: Solve the Resulting Equation
Multiply both sides by \(x - 5\) to cancel the denominator:
[tex]\[ x + 2 = 2(x - 5) \][/tex]
[tex]\[ x + 2 = 2x - 10 \][/tex]
Rearrange to isolate \(x\):
[tex]\[ 2 + 10 = 2x - x \][/tex]
[tex]\[ 12 = x \][/tex]
### Step 6: Verify the Solution
Ensure that the solution makes the original logarithmic arguments valid:
- \( x + 2 \) must be greater than 0: \( 12 + 2 > 0 \) which simplifies to \( 14 > 0 \) (true).
- \( x - 5 \) must be greater than 0: \( 12 - 5 > 0 \) which simplifies to \( 7 > 0 \) (true).
Since both conditions are satisfied, the solution \( x = 12 \) is valid.
### Conclusion
The solution to the equation \(\log_2(x + 2) = 1 + \log_2(x - 5)\) is
[tex]\[ x = 12 \][/tex]
[tex]\[ \log_2(x + 2) = 1 + \log_2(x - 5), \][/tex]
follow these steps:
### Step 1: Understand the Equation
First, recognize that the logarithms are both to the base 2.
[tex]\[ \log_2(x + 2) \][/tex]
[tex]\[ \log_2(x - 5) + 1 \][/tex]
### Step 2: Isolate the Logarithmic Terms
To simplify the equation, subtract \(\log_2(x - 5)\) from both sides:
[tex]\[ \log_2(x + 2) - \log_2(x - 5) = 1 \][/tex]
### Step 3: Apply Logarithmic Properties
Use the property of logarithms that states \(\log_b(a) - \log_b(c) = \log_b\left( \frac{a}{c} \right)\):
[tex]\[ \log_2 \left(\frac{x + 2}{x - 5}\right) = 1 \][/tex]
### Step 4: Eliminate the Logarithm
Remember, if \(\log_b(A) = C\), then \(A = b^C\). In this case, \(b = 2\) and \(C = 1\):
[tex]\[ \frac{x + 2}{x - 5} = 2^1 \][/tex]
[tex]\[ \frac{x + 2}{x - 5} = 2 \][/tex]
### Step 5: Solve the Resulting Equation
Multiply both sides by \(x - 5\) to cancel the denominator:
[tex]\[ x + 2 = 2(x - 5) \][/tex]
[tex]\[ x + 2 = 2x - 10 \][/tex]
Rearrange to isolate \(x\):
[tex]\[ 2 + 10 = 2x - x \][/tex]
[tex]\[ 12 = x \][/tex]
### Step 6: Verify the Solution
Ensure that the solution makes the original logarithmic arguments valid:
- \( x + 2 \) must be greater than 0: \( 12 + 2 > 0 \) which simplifies to \( 14 > 0 \) (true).
- \( x - 5 \) must be greater than 0: \( 12 - 5 > 0 \) which simplifies to \( 7 > 0 \) (true).
Since both conditions are satisfied, the solution \( x = 12 \) is valid.
### Conclusion
The solution to the equation \(\log_2(x + 2) = 1 + \log_2(x - 5)\) is
[tex]\[ x = 12 \][/tex]
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.