Answer :
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]
