Question

How do you solve `ln (x) + ln (x-2) = ln (3x+14)`?

Answer 1

1) Establish the domain
2) Simplify until you have a polynomial (often linear or quadratic) equation
3) Solve the quadratic equation
4) Determine the solutions w. r. t. the domain

Solution: `x = 7`

Explanation:

1) Establishing the domain

First, let's find out the domain for which the logarithmic terms are defined.

As `log_a(x)` is only defined for `x > 0`, you see that you have following restrictions on `x`:

• `x > 0`
• `x - 2 > 0 => x > 2`
• `3x + 14 > 0 => x > - 14/3`

The most restrictive one is `x > 2` since if this condition holds, all the others also hold automatically.

So, any possible solutions need to satisfy `x > 2`.

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2) Simplifying

Now, let's simplify your equation using the logarithmic rule

`log_a(x) + log_a(y) = log_a(x * y)`

In your case, it means:

`ln (x) + ln(x-2) = ln(3x+ 14)`

`<=> ln ( x * (x-2) ) = ln(3x + 14)`

Now we can use that

`log_a(x) = log_a(y) <=> x = y`

for `x > 0`, `y > 0` and `a != 1`. This means that you can drop the `ln` on both sides of the equation which leads to:

`<=> x (x-2) = 3x + 14`

`<=> x^2 - 2x = 3x + 14`

`<=> x^2 - 5x - 14 = 0`

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3) Solving the quadratic equation

At this point, we have a regular quadratic equation which can be solved with different methods. One of the most popular ones that always work is using the quadratic formula

`x = (-b +- sqrt(b^2 - 4ac))/(2a)`

with `a = 1`, `b = -5` and `c = -14`.

Here, the solutions are

`x = 7 color(white)(xx) "or" color(white)(xx) x = -2`

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Alternative method:

Let me show you a different method though that works here too. It is especially easy if `a = 1` and if the solutions are integers.

The trick is to factorize your `x^2 + bx + c` term so that

`x^2 +bx + c = (x + u)(x + v)`

and if you succeed doing so, `x = - u` and `x = - v` (mind the minus!) are your solutions.

So, the goal is finding two integers `u` and `v` so that

`u + v = b` and `u * v = c`

both hold at the same time.

It's easy to see that both equations

`u + v = -5` and `u * v = - 14`

work for

`u = -7` and `v = 2`,

so you can factorize your equation as follows:

`<=> (x - 7)(x + 2) = 0`

`<=> x = 7 color(white)(xx) "or" color(white)(xx) x = -2`

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4) Determining the solution w.r.t. domain

Now, as we have stated that our domain is `x > 2`, we need to discard the second solution `x = -2` since it doesn't fit the condition.

`x = 7` fulfills the condition though since `7 > 2`, so this is the solution of the logarithmic equation.

`color(white)(xx)`

Hope that this helped!