# How do you solve log_4 x^2 - log_4 (x+1) = 5?

Feb 8, 2016

$x \approx - 0.999 \text{ or } x \approx 1024.999$

#### Explanation:

1) Determine the domain

First of all, let's determin the domain of the two logarithmic expressions.

• As ${x}^{2} \ge 0$ for all $x \in \mathbb{R}$, ${\log}_{4} \left({x}^{2}\right)$ is defined for all $x \ne 0$.
• ${\log}_{4} \left(x + 1\right)$ is defined for $x + 1 > 0 \iff x > - 1$

Thus, our domain is $x > - 1$ and $x \ne 0$.

2) Solve the equation

Now let's solve the equation. To start, use the logarithmic law

${\log}_{a} \left(m\right) - {\log}_{a} \left(n\right) = {\log}_{a} \left(\frac{m}{n}\right)$

Thus, you can transform your equation into:

${\log}_{4} \left({x}^{2} / \left(x + 1\right)\right) = 5$

Now, the inverse function of ${\log}_{4} \left(x\right)$ is ${4}^{x}$ which means that both ${\log}_{4} \left({4}^{x}\right) = x$ and ${4}^{{\log}_{4} \left(x\right)} = x$ hold.

Thus, to "eliminate" the logarithmic term, you need to apply ${4}^{x}$ to both sides of the equation:

${x}^{2} / \left(x + 1\right) = {4}^{5}$

${x}^{2} / \left(x + 1\right) = 1024$

... multiply both sides with $\left(x + 1\right)$...

${x}^{2} = 1024 \left(x + 1\right)$

... bring all the terms to the left side...

${x}^{2} - 1024 x - 1024 = 0$

Solve the equation e.g. with the quadratic formula:

$x = \frac{1024 \pm \sqrt{{1024}^{2} + 4 \cdot 1024}}{2} \approx \frac{1024 \pm 1025.998}{2}$

$x \approx - 0.999 \text{ or } x \approx 1024.999$

3) Check the domain

Now, we need to check if both $x$ fit into our domain.

Indeed, $x > - 1$ and $x \ne 0$ holds for both of them, so both are solutions.