Christmas Streak 35/88: Seems Easy Enough!

Relevant wiki: Properties of Logarithms - Basic

Using the basic rules of logarithms, we can simplify the equation to $\displaystyle \log_{x-1} \left\{\dfrac{1}{6}(-x+4)(-x^3+7x^2-13x+9)\right\}=0.$

Since the antilogarithm must equal 1, we can write $(x-4)(x^3-7x^2+13x-9)=6.$

Simplifying this, we get $(x-1)(x-2)(x-3)(x-5)=0.$

Then are there 4 values for $x$ ?

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Woah, wait, no, hold on there for a sec. Calm down.

We're taking the real logarithm. Check if the base is positive and not equal to 1, and also if the antilogarithm isn't negative.

If $x=1,$ the base is equal to $0.$ Then all the logs aren't defined.

If $x=2,$ the base is equal to $1.$ Similarly, all the logs aren't defined.

If $x=3,$ the equation would look like: $\displaystyle \log_{2} \frac{1}{6}+\log_{2}1+\log_{2}6=0.$ Makes perfect sense.

If $x=5,$ the second and third antilogarithm is negative. Then those two logs aren't defined.

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From above, we can figure out that there is only 1 value for $x,$ which is $3.$

(P.S. If you type this up in wolframalpha, it'll tell you x = 1 or 3. Defeating the computer is easy. >w>)