Math Problem Statement

Log (x - 1) + log (x + 1) = 1, find the value of x.

Solution

We are given the equation:

log(x1)+log(x+1)=1\log(x - 1) + \log(x + 1) = 1

Step 1: Use the logarithmic property

We can combine the logarithms using the property of logarithms:

log(a)+log(b)=log(ab)\log(a) + \log(b) = \log(a \cdot b)

So, applying this to the given equation:

log((x1)(x+1))=1\log((x - 1)(x + 1)) = 1

Step 2: Simplify the expression

The product (x1)(x+1)(x - 1)(x + 1) simplifies to:

(x1)(x+1)=x21(x - 1)(x + 1) = x^2 - 1

Now, we have:

log(x21)=1\log(x^2 - 1) = 1

Step 3: Convert from logarithmic form to exponential form

Recall that logb(y)=z\log_b(y) = z implies y=bzy = b^z. Here, since we are using the common logarithm (base 10), we convert the equation to:

x21=101x^2 - 1 = 10^1

x21=10x^2 - 1 = 10

Step 4: Solve for xx

Now, solve the equation:

x2=10+1=11x^2 = 10 + 1 = 11

x=±11x = \pm \sqrt{11}

Step 5: Check for validity

Since the logarithmic function is defined only for positive arguments, we must ensure that both x1>0x - 1 > 0 and x+1>0x + 1 > 0. This implies:

x1>0orx>1x - 1 > 0 \quad \text{or} \quad x > 1

Therefore, x=11x = \sqrt{11} is the only valid solution (as x=11x = -\sqrt{11} would make log(x1)\log(x - 1) undefined).

Final Answer:

The value of xx is:

x=11x = \sqrt{11}

Would you like more details on any step?

Here are 5 related questions to practice:

  1. Solve log(x+3)+log(x2)=2\log(x + 3) + \log(x - 2) = 2.
  2. Solve log(2x)+log(3)=1\log(2x) + \log(3) = 1.
  3. If log(x24)=0\log(x^2 - 4) = 0, find xx.
  4. Solve log(x+5)log(x1)=1\log(x + 5) - \log(x - 1) = 1.
  5. Find xx if log(x29)=1\log(x^2 - 9) = 1.

Tip: Always check the domain restrictions when solving logarithmic equations, as logarithms are only defined for positive values.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Algebra
Exponential Equations

Formulas

log(a) + log(b) = log(a * b)
log_b(y) = z implies y = b^z

Theorems

Logarithmic properties
Exponential form of logarithmic equations

Suitable Grade Level

Grades 9-12