Solving for $x$ in this equation

Question

Solve for $x$:

$$\{x\mid (3x)^2 = 30 \cdot 3^x - 81\}$$

Edit: I have copied the above question word for word from a grade-10 (high-school) textbook. I believe that there may possibly be a printing error as the answer in the textbook states $x$ is 1,3. From the answers being given below it is evident that the kind of solution being given is way above any of the concepts grade 10 students are learning. Moderators may delete this question if they feel it appropriate. I apologise for any time wasted. Kind regards.

Answer 1

HINT

• You can simplify to $(3x)^2 = 30 \cdot 3^x - 81\implies 3x^2=10\cdot 3^x-27$
• Let study the roots of $f(x)=3x^2-10\cdot 3^x+27$ (number of roots and approximate values)
• For the exact solution you need numerical method

Answer 2

Consider $f(x) = 30\cdot 3^x - 3x^2 - 81$. Best way is to divide $\mathbb{R}$ into three parts. So if $x \ge 1$, $f'(x) = 30\cdot 3^x\cdot \log 3 - 6x > 6(3^x-x) > 0\implies f(x) \ge f(1) = 90 - 3-81 = 6 > 0$. If $x \le 0\implies f'(x) > 0\implies f(x) \le f(0) = -81 < 0$. If $0 < x < 1\implies f'(x) > 30\cdot \log 3 - 6 > 0 $, and also $f(0) = -51 < 0, f(1) = 6 > 0$ and since $f$ is continuous , IVT says there is a unique solution of $f(x) = 0$ on $(0,1)$ which can be found by Lambert function and by Wolfram Alpha or by Newton method of approximation.