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Solve: log4 (x2 – 9) – log4 (x + 3) = log4 64.
The Correct Answer and Explanation is:
We are given the logarithmic equation: log4(x2−9)−log4(x+3)=log464\log_4 (x^2 – 9) – \log_4 (x + 3) = \log_4 64
Step 1: Apply Logarithmic Rules
Using the logarithmic identity: logbA−logbB=logb(AB)\log_b A – \log_b B = \log_b \left(\frac{A}{B}\right)
We simplify the left-hand side: log4(x2−9x+3)=log464\log_4 \left( \frac{x^2 – 9}{x + 3} \right) = \log_4 64
Step 2: Eliminate the Logarithms
Since both sides have the same logarithmic base, we can equate the arguments: x2−9x+3=64\frac{x^2 – 9}{x + 3} = 64
Step 3: Simplify the Left Side
Notice that x2−9x^2 – 9 is a difference of squares: x2−9=(x−3)(x+3)x^2 – 9 = (x – 3)(x + 3)
So, (x−3)(x+3)x+3\frac{(x – 3)(x + 3)}{x + 3}
As long as x≠−3x \neq -3, the x+3x + 3 terms cancel: x−3=64x – 3 = 64
Step 4: Solve for xx
x=64+3=67x = 64 + 3 = 67
Step 5: Check for Restrictions
We must check that the values inside the logarithms are positive:
- x2−9=672−9=4489−9=4480>0x^2 – 9 = 67^2 – 9 = 4489 – 9 = 4480 > 0
- x+3=67+3=70>0x + 3 = 67 + 3 = 70 > 0
So, x=67x = 67 is a valid solution.
✅ Final Answer:
67\boxed{67}
Explanation
This logarithmic equation tests your understanding of logarithmic properties and algebraic manipulation. The original equation contains a difference of two logarithms on the left and a single logarithm on the right. Because all the logarithms share the same base (base 4), we can simplify the expression by using a logarithmic identity: logbA−logbB=logb(AB)\log_b A – \log_b B = \log_b \left(\frac{A}{B}\right)
This rule allows us to combine the two logarithms into a single one: log4(x2−9x+3)\log_4 \left(\frac{x^2 – 9}{x + 3}\right). Once we simplify the rational expression inside the log, we notice that the numerator is a difference of squares. Factoring it gives (x−3)(x+3)(x – 3)(x + 3). When we divide this by x+3x + 3, the x+3x + 3 terms cancel out (as long as x≠−3x \neq -3, which would make the denominator zero), and we are left with x−3x – 3.
Now, we have log4(x−3)=log464\log_4 (x – 3) = \log_4 64, so we equate the arguments and solve x−3=64x – 3 = 64, leading to x=67x = 67.
We must check that this value doesn’t make any part of the logarithm undefined (logarithms are only defined for positive arguments). Since all values involved are positive when x=67x = 67, the solution is valid.
Thus, the correct and verified solution is 67\boxed{67}.
