Question

4 Log (x+1) = 2 Log (x+3)​

Answer 1

The value of x is 1.

To solve the equation
`\(4 \log(x+1) = 2 \log(x+3)\)`, we can use logarithmic properties. Specifically, we can use the fact that
`\(\log(a^n) = n \log(a)\)`.

The equation becomes:

`\[ \log((x+1)^4) = \log((x+3)^2) \]`

Now, we can remove the logarithm by setting the expressions inside the logs equal to each other:

`\[ (x+1)^4 = (x+3)^2 \]`

Expand both sides:

`\[ (x+1)(x+1)(x+1)(x+1) = (x+3)(x+3) \]`

Simplify further:

`\[ (x+1)^2(x+1)^2 = (x+3)^2 \]`

Now, substitute
`\(y = (x+1)^2\)`:

`\[ y^2 = (x+3)^2 \]`

Take the square root of both sides:

`\[ y = x+3 \]`

Substitute back
`\(y = (x+1)^2\)`:

`\[ (x+1)^2 = x+3 \]`

Expand and simplify:

`\[ x^2 + 2x + 1 = x + 3 \]`

Subtract
`\(x\)` from both sides:

`\[ x^2 + x - 2 = 0 \]`

Now, factor the quadratic equation:

`\[ (x - 1)(x + 2) = 0 \]`

So, x = 1 or x = -2. However, we need to check if these solutions are valid for the original equation.

Let's check x = 1:

`\[4 \log(1+1) = 4 \log(2)\]`

`\[2 \log(1+3) = 2 \log(4)\]`

Since
`\(4 \log(2) = 2 \log(4)\)`, x = 1 is a valid solution.

Now, check x = -2:

`\[4 \log(-2+1)\]` is not defined, so
`\(x = -2\)` is not a valid solution.

Therefore, the only valid solution is x = 1.

The probable question may be:

"Solve for x in: 4 Log (x+1) = 2 Log (x+3)​ "