Question

Solve the logarithmic equation%3D%202" id="TexFormula1" title=" log_{6}(2x - 6) + log_{6}x = 2" alt=" log_{6}(2x - 6) + log_{6}x = 2" align="absmiddle" class="latex-formula">
​

Answer 1

Answer:

$x=6$

Step-by-step explanation:

We want to solve the equation:

$\displaystyle \log_6(2x-6)+\log_6x=2$

Recall the property:

$\displaystyle \log_bx+\log_by=\log_b(xy)$

Hence:

$\log_6(x(2x-6))=2$

Next, recall that by the definition of logarithms:

$\displaystyle \log_b(a)=c\text{ if and only if } b^c=a$

Therefore:

$6^2=x(2x-6)$

Solve for x. Simplify and distribute:

$36=2x^2-6x$

We can divide both sides by two:

$x^2-3x=18$

Subtract 18 from both sides:

$x^2-3x-18=0$

Factor:

$(x-6)(x+3)=0$

Zero Product Property:

$x-6=0\text{ or } x+3=0$

Solve for each case. Hence:

$x=6\text{ or } x=-3$

Next, we must check the solutions for extraneous solutions. To do so, we can simply substitute the solutions back into the original equations and examine its validity.

Checking x = 6:

$\displaystyle \begin{aligned} \log_{6}(2(6)-6)+\log_{6}6&\stackrel{?}{=} 2 \\ \\ \log_6(12-6)+(1)&\stackrel{?}{=}2 \\ \\ \log_6(6)+1&\stackrel{?}{=}2 \\ \\ 1+1=2&\stackrel{\checkmark}{=}2\end{aligned}$

Hence, x = 6 is indeed a solution.

Checking x = -3:

$\displaystyle\begin{aligned} \log_6(2(-3)-6) + \underbrace{\log_6-3}_{\text{und.}} &\stackrel{?}{=} 2\\ \\ \end{aligned}$

Since the second term is undefined, x = -3 is not a solution.

Therefore, our only solution is x = 6.

Answer 2

Answer:

x = 6

Step-by-step explanation:

The given logarithmic equation is ,

$\implies log_{6}(2x - 6) + log_{6}x = 2$

We can notice that the bases of both logarithm is same . So we can use a property of log as ,

$\bf \to log_a b + log_a c = log_a {( ac)}$

So we can simplify the LHS and write it as ,

$\implies log_{6} \{ x ( 2x - 6 )\} = 2$

Now simplify out x(2x - 6 ) . We get ,

$\implies log_6 ( 2x^2 - 6x ) = 2$

Again , we know that ,

$\bf \to log_a b = c , a^c = b$

Using this we have ,

$\implies 2x^2 - 6x = 6^2 \\\\\implies 2x^2 - 6x -36 = 0$

Now simplify the quadratic equation ,

$\implies x^2 - 3x - 18 = 0 \\\\\implies x^2 -6x + 3x -18=0\\\\\implies x( x -6) +3( x - 6 ) = 0 \\\\\implies (x-6)(x+3) = 0 \\\\\implies x = 6 , -3$

Since logarithms are not defined for negative numbers or zero , therefore ,

$\implies 2x - 6 > 0 \\\\\implies x > 3$

Therefore the equation is not defined at x = -3 . Hence the possible value of x is 6 .

$\implies \underline{\underline{ x \quad = \quad 6 }}$