Notes: `log_4(-x)+log_4(x+10)=2` Solve the equation. Check for extraneous solutions.

Monday, 28 March 2016

$\displaystyle{\log}_{{4}}{\left(-{x}\right)}+{\log}_{{4}}{\left({x}+{10}\right)}={2}$ Solve the equation. Check for extraneous solutions.

To evaluate the given equation $\displaystyle{\log}_{{4}}{\left(-{x}\right)}+{\log}_{{4}}{\left({x}+{10}\right)}={2}$ , we may apply the logarithm property: $\displaystyle{\log}_{{b}}{\left({x}\right)}+{\log}_{{b}}{\left({y}\right)}={\log}_{{b}}{\left({x}\cdot{y}\right)}$ .

$\displaystyle{\log}_{{4}}{\left(-{x}\right)}+{\log}_{{4}}{\left({x}+{10}\right)}={2}$

$\displaystyle{\log}_{{4}}{\left({\left(-{x}\right)}\cdot{\left({x}+{10}\right)}\right)}={2}$

$\displaystyle{\log}_{{4}}{\left(-{{x}}^{{2}}-{10}{x}\right)}={2}$

To get rid of the "log" function, we may apply the logarithm property: $\displaystyle{{b}}^{{{\log}_{{b}}{\left({x}\right)}}}={x}.$

Raise both sides by base of $\displaystyle{4}$ .

$\displaystyle{{4}}^{{{\log}_{{4}}{\left(-{{x}}^{{2}}-{10}{x}\right)}}}={{4}}^{{2}}$

$\displaystyle-{{x}}^{{2}}-{10}{x}={16}$

Add $\displaystyle{{x}}^{{2}}$ and $\displaystyle{10}{x}$ on both sides of the equation to simplify in standard form: $\displaystyle{a}{{x}}^{{2}}+{b}{x}+{c}={0}.$

$\displaystyle-{{x}}^{{2}}-{10}{x}+{{x}}^{{2}}+{10}{x}={16}+{{x}}^{{2}}+{10}{x}$

$\displaystyle{0}={16}+{{x}}^{{2}}+{10}{x}{\quad\text{or}\quad}{{x}}^{{2}}+{10}{x}+{16}={0}.$

Apply factoring on the trinomial.

$\displaystyle{\left({x}+{2}\right)}{\left({x}+{8}\right)}={0}$

Apply zero-factor property to solve for x...

$\displaystyle{\log}_{{4}}{\left(-{x}\right)}+{\log}_{{4}}{\left({x}+{10}\right)}={2}$

$\displaystyle{\log}_{{4}}{\left({\left(-{x}\right)}\cdot{\left({x}+{10}\right)}\right)}={2}$

$\displaystyle{\log}_{{4}}{\left(-{{x}}^{{2}}-{10}{x}\right)}={2}$

To get rid of the "log" function, we may apply the logarithm property: $\displaystyle{{b}}^{{{\log}_{{b}}{\left({x}\right)}}}={x}.$

Raise both sides by base of $\displaystyle{4}$ .

$\displaystyle{{4}}^{{{\log}_{{4}}{\left(-{{x}}^{{2}}-{10}{x}\right)}}}={{4}}^{{2}}$

$\displaystyle-{{x}}^{{2}}-{10}{x}={16}$

Add $\displaystyle{{x}}^{{2}}$ and $\displaystyle{10}{x}$ on both sides of the equation to simplify in standard form: $\displaystyle{a}{{x}}^{{2}}+{b}{x}+{c}={0}.$

$\displaystyle-{{x}}^{{2}}-{10}{x}+{{x}}^{{2}}+{10}{x}={16}+{{x}}^{{2}}+{10}{x}$

$\displaystyle{0}={16}+{{x}}^{{2}}+{10}{x}{\quad\text{or}\quad}{{x}}^{{2}}+{10}{x}+{16}={0}.$

Apply factoring on the trinomial.

$\displaystyle{\left({x}+{2}\right)}{\left({x}+{8}\right)}={0}$

Apply zero-factor property to solve for x by equating each factor to $\displaystyle{0}$ .

$\displaystyle{x}+{2}={0}$

$\displaystyle{x}+{2}-{2}={0}-{2}$

$\displaystyle{x}=-{2}$

and

$\displaystyle{x}+{8}={0}$

$\displaystyle{x}+{8}-{8}={0}-{8}$

$\displaystyle{x}=-{8}$

Checking: Plug-in each $\displaystyle{x}$ on $\displaystyle{\log}_{{4}}{\left(-{x}\right)}+{\log}_{{4}}{\left({x}+{10}\right)}={2}$ .

Let $\displaystyle{x}=-{2}$ on $\displaystyle{\log}_{{4}}{\left(-{x}\right)}+{\log}_{{4}}{\left({x}+{10}\right)}={2}$ .

$\displaystyle{\log}_{{4}}{\left(-{\left(-{2}\right)}\right)}+{\log}_{{4}}{\left(-{2}+{10}\right)}=?{2}$

$\displaystyle{\log}_{{4}}{\left({2}\right)}+{\log}_{{4}}{\left({8}\right)}=?{2}$

$\displaystyle{\log}_{{4}}{\left({2}\cdot{8}\right)}=?{2}$

$\displaystyle{\log}_{{4}}{\left({16}\right)}=?{2}$

$\displaystyle{\log}_{{4}}{\left({{4}}^{{2}}\right)}=?{2}$

$\displaystyle{2}{\log}_{{4}}{\left({4}\right)}=?{2}$

$\displaystyle{2}\cdot{1}=?{2}$

$\displaystyle{2}={2}$ TRUE

Let $\displaystyle{x}=-{8}$ on $\displaystyle{\log}_{{4}}{\left(-{x}\right)}+{\log}_{{4}}{\left({x}+{10}\right)}={2}.$

$\displaystyle{\log}_{{4}}{\left(-{\left(-{8}\right)}\right)}+{\log}_{{4}}{\left(-{8}+{10}\right)}=?{2}$

$\displaystyle{\log}_{{4}}{\left({8}\right)}+{\log}_{{4}}{\left({2}\right)}=?{2}$

$\displaystyle{\log}_{{4}}{\left({8}\cdot{2}\right)}=?{2}$

$\displaystyle{\log}_{{4}}{\left({16}\right)}=?{2}$

$\displaystyle{2}={2}$ TRUE

Therefore, there are no extraneous solutions.

Both solved x-values: $\displaystyle{x}=-{2}$ and $\displaystyle{x}=-{8}$ are real solution of the equation $\displaystyle{\log}_{{4}}{\left(-{x}\right)}+{\log}_{{4}}{\left({x}+{10}\right)}={2}$ .