Let's solve the given equation:

$\log(x) + \log(x+1) = 1$

Step 1: Apply Logarithm Properties

We can combine the two logarithms using the property:

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

Applying this to the equation:

$\log(x) + \log(x+1) = \log(x \cdot (x+1)) = 1$

So now we have:

$\log(x(x+1)) = 1$

Step 2: Convert to Exponential Form

To solve for $x$, we need to rewrite the logarithmic equation in its exponential form. Since the logarithm base is not specified, we assume it is base 10:

$x(x+1) = 10^1$

$x(x+1) = 10$

Step 3: Expand and Rearrange

Expanding the product gives:

$x^2 + x = 10$

Subtract 10 from both sides to set the equation to zero:

$x^2 + x - 10 = 0$

Step 4: Solve the Quadratic Equation

We can solve the quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For our equation $x^2 + x - 10 = 0$, $a = 1$, $b = 1$, and $c = -10$. Plugging in these values:

$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1}$

$x = \frac{-1 \pm \sqrt{1 + 40}}{2}$

$x = \frac{-1 \pm \sqrt{41}}{2}$

Step 5: Determine the Possible Solutions

We have two possible solutions:

$x = \frac{-1 + \sqrt{41}}{2} \quad \text{or} \quad x = \frac{-1 - \sqrt{41}}{2}$

Since $x$ must be positive (logarithms are only defined for positive values of $x$), we discard the negative solution. Therefore:

$x = \frac{-1 + \sqrt{41}}{2}$

Step 6: Approximate the Solution

To approximate:

$x \approx \frac{-1 + 6.403}{2} \approx \frac{5.403}{2} \approx 2.7015$

Final Answer

$x \approx 2.7015$

Would you like more details or have any questions about this solution?

Here are some related questions to further your understanding:

1. How would the solution change if the base of the logarithms were different?
2. What are the properties of logarithms that can help simplify expressions?
3. How can you solve quadratic equations if the quadratic formula isn’t an option?
4. What are other methods to solve equations involving logarithms?
5. Can you solve a similar logarithmic equation involving natural logs ($\ln$)?

Tip: Always remember to check the domain when solving logarithmic equations, as the argument of a logarithm must be positive.