Notes: `log_6(3x)+log_6(x-1)=3` Solve the equation. Check for extraneous solutions.

Wednesday, 3 May 2017

$\displaystyle{\log}_{{6}}{\left({3}{x}\right)}+{\log}_{{6}}{\left({x}-{1}\right)}={3}$ Solve the equation. Check for extraneous solutions.

To evaluate the given equation $\displaystyle{\log}_{{6}}{\left({3}{x}\right)}+{\log}_{{6}}{\left({x}-{1}\right)}={3}$ , we may apply the logarithm property: $\displaystyle{\log}_{{b}}{\left({x}\right)}+{\log}_{{b}}{\left({y}\right)}={\log}_{{b}}{\left({x}\cdot{y}\right)}$ .

$\displaystyle{\log}_{{6}}{\left({3}{x}\right)}+{\log}_{{6}}{\left({x}-{1}\right)}={3}$

$\displaystyle{\log}_{{6}}{\left({3}{x}\cdot{\left({x}-{1}\right)}\right)}={3}$

$\displaystyle{\log}_{{6}}{\left({3}{{x}}^{{2}}-{3}{x}\right)}={3}$

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 $\displaystyle{6}$ .

$\displaystyle{{6}}^{{{\log}_{{6}}{\left({3}{{x}}^{{2}}-{3}{x}\right)}}}={{6}}^{{3}}$

$\displaystyle{3}{{x}}^{{2}}-{3}{x}={216}$

Subtract $\displaystyle{216}$ from both sides of the equation to simplify in standard form: $\displaystyle{a}{{x}}^{{2}}+{b}{x}+{c}={0}$ .

$\displaystyle{3}{{x}}^{{2}}-{3}{x}-{216}={216}-{216}$

$\displaystyle{3}{{x}}^{{2}}-{3}{x}-{216}={0}$

Apply factoring on the trinomial.

$\displaystyle{3}\cdot{\left({x}+{8}\right)}\cdot{\left({x}-{9}\right)}={0}$

Apply zero-factor property to solve for...

$\displaystyle{\log}_{{6}}{\left({3}{x}\right)}+{\log}_{{6}}{\left({x}-{1}\right)}={3}$

$\displaystyle{\log}_{{6}}{\left({3}{x}\cdot{\left({x}-{1}\right)}\right)}={3}$

$\displaystyle{\log}_{{6}}{\left({3}{{x}}^{{2}}-{3}{x}\right)}={3}$

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 $\displaystyle{6}$ .

$\displaystyle{{6}}^{{{\log}_{{6}}{\left({3}{{x}}^{{2}}-{3}{x}\right)}}}={{6}}^{{3}}$

$\displaystyle{3}{{x}}^{{2}}-{3}{x}={216}$

Subtract $\displaystyle{216}$ from both sides of the equation to simplify in standard form: $\displaystyle{a}{{x}}^{{2}}+{b}{x}+{c}={0}$ .

$\displaystyle{3}{{x}}^{{2}}-{3}{x}-{216}={216}-{216}$

$\displaystyle{3}{{x}}^{{2}}-{3}{x}-{216}={0}$

Apply factoring on the trinomial.

$\displaystyle{3}\cdot{\left({x}+{8}\right)}\cdot{\left({x}-{9}\right)}={0}$

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

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

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

$\displaystyle{x}-{9}={0}$

$\displaystyle{x}-{9}+{9}={0}+{9}$

$\displaystyle{x}={9}$

Checking: Plug-in each $\displaystyle{x}$ on $\displaystyle{\log}_{{6}}{\left({3}{x}\right)}+{\log}_{{6}}{\left({x}-{1}\right)}={3}$ .

Let $\displaystyle{x}=-{8}$ on $\displaystyle{\log}_{{6}}{\left({3}{x}\right)}+{\log}_{{6}}{\left({x}-{1}\right)}={3}$ .

$\displaystyle{\log}_{{6}}{\left({3}\cdot{\left(-{8}\right)}\right)}+{\log}_{{6}}{\left(-{8}-{1}\right)}=?{3}$

$\displaystyle{\log}_{{6}}{\left(-{24}\right)}+{\log}_{{6}}{\left(-{9}\right)}=?{3}$

undefined +undefined =?3 FALSE

Note that $\displaystyle{\log}_{{b}}{\left({x}\right)}$ is undefined on $\displaystyle{x}\leq{0}$ .

Let $\displaystyle{x}={9}$ on $\displaystyle{\log}_{{6}}{\left({3}{x}\right)}+{\log}_{{6}}{\left({x}-{1}\right)}={3}$ .

$\displaystyle{\log}_{{6}}{\left({3}\cdot{9}\right)}+{\log}_{{6}}{\left({9}-{1}\right)}=?{3}$

$\displaystyle{\log}_{{6}}{\left({27}\right)}+{\log}_{{6}}{\left({8}\right)}=?{3}$

$\displaystyle{\log}_{{6}}{\left({27}\cdot{8}\right)}=?{3}$

$\displaystyle{\log}_{{6}}{\left({216}\right)}=?{3}$

$\displaystyle{\log}_{{6}}{\left({{6}}^{{3}}\right)}=?{3}$

$\displaystyle{3}{\log}_{{6}}{\left({6}\right)}=?{3}$

$\displaystyle{3}\cdot{1}=?{3}$

$\displaystyle{3}={3}$ TRUE

Thus, the $\displaystyle{x}=-{8}$ is an extraneous solution.

The $\displaystyle{x}={9}$ is the only real solution of the equation $\displaystyle{\log}_{{6}}{\left({3}{x}\right)}+{\log}_{{6}}{\left({x}-{1}\right)}={3}$ .