## solve the logarithmic equation $$log_{6}(2x – 6) + log_{6}x = 2$$ ​

Question

solve the logarithmic equation
$$log_{6}(2x – 6) + log_{6}x = 2$$

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1 year 2021-08-01T22:51:04+00:00 2 Answers 5 views 0

$$x=6$$

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.

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 .

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