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

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

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Answer 1 · 2015-11-30T13:41:30

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: $x = 7$ $x = 7$

Explanation:

1) Establishing the domain

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

As ${log}_{a} (x)$ $log_a(x)$ is only defined for $x > 0$ $x > 0$ , you see that you have following restrictions on $x$ $x$ :

$x > 0$ $x > 0$
$x - 2 > 0 \Rightarrow x > 2$ $x - 2 > 0 => x > 2$
$3 x + 14 > 0 \Rightarrow x > - \frac{14}{3}$ $3x + 14 > 0 => x > - 14/3$

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

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

============================================

2) Simplifying

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

${log}_{a} (x) + {log}_{a} (y) = {log}_{a} (x \cdot y)$ $log_a(x) + log_a(y) = log_a(x * y)$

In your case, it means:

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

$\Leftrightarrow ln (x \cdot (x - 2)) = ln (3 x + 14)$ $<=> ln ( x * (x-2) ) = ln(3x + 14)$

Now we can use that

${log}_{a} (x) = {log}_{a} (y) \Leftrightarrow x = y$ $log_a(x) = log_a(y) <=> x = y$

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

$\Leftrightarrow x (x - 2) = 3 x + 14$ $<=> x (x-2) = 3x + 14$

$\Leftrightarrow x^{2} - 2 x = 3 x + 14$ $<=> x^2 - 2x = 3x + 14$

$\Leftrightarrow x^{2} - 5 x - 14 = 0$ $<=> x^2 - 5x - 14 = 0$

============================================

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

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2 - 4ac))/(2a)$

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

Here, the solutions are

$x = 7 \times or \times x = - 2$ $x = 7 color(white)(xx) "or" color(white)(xx) x = -2$

========

Alternative method:

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

The trick is to factorize your $x^{2} + b x + c$ $x^2 + bx + c$ term so that

$x^{2} + b x + c = (x + u) (x + v)$ $x^2 +bx + c = (x + u)(x + v)$

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

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

$u + v = b$ $u + v = b$ and $u \cdot v = c$ $u * v = c$

both hold at the same time.

It's easy to see that both equations

$u + v = - 5$ $u + v = -5$ and $u \cdot v = - 14$ $u * v = - 14$

work for

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

so you can factorize your equation as follows:

$\Leftrightarrow (x - 7) (x + 2) = 0$ $<=> (x - 7)(x + 2) = 0$

$\Leftrightarrow x = 7 \times or \times x = - 2$ $<=> x = 7 color(white)(xx) "or" color(white)(xx) x = -2$

============================================

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

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

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

$\times$ $color(white)(xx)$

Hope that this helped!

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