Algebra Problem of The Day-6

[BigBeachBanana]

$$\large x-6 +\frac{2}{\sqrt{x-2}}=\frac{1}{3}\log_3\!\left({\frac{x}{x^3+54}}\right)$$
Solve for $\large x\in \R.$

 

[Steven G]

Oh no, here we go again!

 

[BigBeachBanana]

Oh no, here we go again!
Should the 54 be 64?

Yes, let's go again!
54 is correct. You can cheat a bit by graphing the equation to see what the answer is supposed to be. It might give you an idea of how to approach it. I had to be very creative for this one.

Looking forward to your solution.

 

[Steven G]

Spoiler

With 3^3 + 54 =81 = 3^4, I immediately plugged in x=3 and it worked! Now I have to show my work. Give me a few minutes!

 

[Steven G]

Yes, let's go again!
54 is correct. You can cheat a bit by graphing the equation to see what the answer is supposed to be. It might give you an idea of how to approach it. I had to be very creative for this one.

Looking forward to your solution.

I already have an answer. Is that quick enough for you.

 

[Steven G]

I'm ready for a hint.

 

[BigBeachBanana]

I'm ready for a hint.

Spoiler

let $3^y=\frac{x}{x^3+54}$

 

[red12dog34]

Let $$f(x)=x-6+\frac{2}{\sqrt{x-2}}, \ g(x)=\frac{1}{3\ln 3}\ln\left(\frac{x}{x^3+54}\right), \ x>2$$Then $$f'(x)=1-\frac{1}{(x-2)\sqrt{x-2}}$$$$f'(x)=0\Rightarrow x=3, \ f(3)=-1$$$$f'(x)<0, \ \forall x\in(2,3), \ f'(x)>0, \ \forall x\in(3,\infty)\Rightarrow f(x)\ge -1, \ \forall x>2$$$$g'(x)=\frac{27-x^3}{x(x^3+54)\ln 3}$$$$g'(x)=0\Rightarrow x=3, \ g(3)=-1$$$$g'(x)>0, \ \forall x\in(2,3), \ g'(x)<0, \ \forall x\in(3,\infty)\Rightarrow g(x)\le -1, \ \forall x>2$$So $$g(x)\le -1\le f(x), \ \forall x>2$$equality stands only for $$x=3$$

 

[Steven G]

BBB,
If the graph I looked at looked like the one above I might have gotten the answer. Oh, well! I have another chance at today's questions.

 

[lookagain]

Let ...

red12dog34, please hide all future solution posts using the spoiler feature for these puzzle challenges as a courtesy to others looking through the respective threads.