Question

If \[ \log_2(x-1)+\log_2(x+1)=3, \] then \(x\) is equal to:

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(5\)

Hint:

After solving any logarithmic equation, always check whether the obtained solutions make every logarithmic argument positive. This step prevents acceptance of extraneous roots.

Answer 1

The Correct Option is B

Step 1: Combine the logs.
Using $\log_a m+\log_a n=\log_a(mn)$, the left side becomes $\log_2[(x-1)(x+1)]$. So the equation is $\log_2[(x-1)(x+1)]=3$.

Step 2: Simplify the product.
Note $(x-1)(x+1)=x^2-1$. So $\log_2(x^2-1)=3$.

Step 3: Convert to exponential form.
A log equation $\log_2(A)=3$ means $A=2^3$. So $x^2-1=8$.

Step 4: Solve the quadratic.
Add $1$ to both sides: $x^2=9$, so $x=3$ or $x=-3$.

Step 5: Check the domain.
Logs need positive insides. We need $x-1>0$ and $x+1>0$, so $x>1$.

Step 6: Reject the bad root.
The value $x=-3$ fails $x>1$, so we throw it out. Only $x=3$ survives.

Step 7: State the answer.
\[ \boxed{3} \]