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

Nov 30, 2015

1) Establish the domain
2) Simplify until you have a polynomial (often linear or 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 ${\log}_{a} \left(x\right)$ is only defined for $x > 0$, you see that you have following restrictions on $x$:

• $x > 0$
• $x - 2 > 0 \implies x > 2$
• $3 x + 14 > 0 \implies x > - \frac{14}{3}$

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

${\log}_{a} \left(x\right) + {\log}_{a} \left(y\right) = {\log}_{a} \left(x \cdot y\right)$

$\ln \left(x\right) + \ln \left(x - 2\right) = \ln \left(3 x + 14\right)$

$\iff \ln \left(x \cdot \left(x - 2\right)\right) = \ln \left(3 x + 14\right)$

Now we can use that

${\log}_{a} \left(x\right) = {\log}_{a} \left(y\right) \iff x = y$

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

$\iff x \left(x - 2\right) = 3 x + 14$

$\iff {x}^{2} - 2 x = 3 x + 14$

$\iff {x}^{2} - 5 x - 14 = 0$

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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

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

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

Here, the solutions are

$x = 7 \textcolor{w h i t e}{\times} \text{or} \textcolor{w h i t e}{\times} x = - 2$

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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 ${x}^{2} + b x + c$ term so that

${x}^{2} + b x + c = \left(x + u\right) \left(x + v\right)$

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 $u \cdot v = c$

both hold at the same time.

It's easy to see that both equations

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

work for

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

so you can factorize your equation as follows:

$\iff \left(x - 7\right) \left(x + 2\right) = 0$

$\iff x = 7 \textcolor{w h i t e}{\times} \text{or} \textcolor{w h i t e}{\times} x = - 2$

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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.

$\textcolor{w h i t e}{\times}$

Hope that this helped!