Solve for x log base 2 of x+3+ log base 2 of x-3=4

Problem

(log_2)(x+3)+(log_2)(x−3)=4

Solution

1. Apply the product rule for logarithms, which states that (log_b)(M)+(log_b)(N)=(log_b)(M*N)

(log_2)((x+3)*(x−3))=4

2. Simplify the product inside the logarithm using the difference of squares formula (a+b)*(a−b)=a2−b2

(log_2)(x2−9)=4

3. Rewrite in exponential form by using the definition (log_b)(y)=x⇔bx=y

x2−9=2

4. Evaluate the power on the right side of the equation.

x2−9=16

5. Isolate the variable by adding 9 to both sides.

x2=25

6. Solve for x by taking the square root of both sides.

x=±5

7. Check for extraneous solutions by ensuring the arguments of the original logarithms are positive.

x+3>0⇒x>−3

x−3>0⇒x>3

Since x=−5 does not satisfy x>3 we discard it.

Final Answer

x=5

Want more problems? Check here!