solve for $x$ without using softwares $\log_{\sqrt{x}}2+\log_6x^x=4$

Question

Is there any nice way to solve this equation without wolfram?

$\log_{\sqrt{x}}2+\log_6x^x=4$

Thanks.

Answer 1

I am pretty sure the isn't solvable with Algebra II, unless you resort to the mess that is complex logs. However: $$ log_\sqrt{x}2+log_6x^x=4\Rightarrow log_\sqrt{x}6+x(log_26)(log_6x)=4(log_26)\Rightarrow \frac{2}{log_6x}+x(log_26)(log_6x)=4(log_26)\Rightarrow2+x(log_6x)^2-4(log_6x)(log_26)=0 $$ which is a quadratic in $log_6x$ except for the pesky x coefficient. Using the quadratic formula anyway, and evaluating the logs, $$ log_6x\approx\frac{10.34\pm\sqrt{106.91-8x}}{2x} $$ This is definitely solvable. However, The only way I can come up with a numerical solution is graphing the two sides.

Answer 2

Rewriting $$\log_{\sqrt{x}}2+\log_6x^x=4$$ leads to the equation $$f(x)=x \log ^2(x)-4 \log (6) \log (x)+\log (4) \log (6)=0$$ which cannot be solved analytically ($f'(x)=0$ or $f''(x)=0$ cannot be solved either).

Plotting the function shows two roots, the first close to $x=1.5$, the second close to $x=4.0$. For the solution, Newton is the simplest root-finder. Strating from a "reasonable" guess $x_0$, it will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

For the first root, using $x_0=1.5$, the successive iterates are $1.45386$, $1.45521$ which is the solution for six significant figures. Limiting to a single iteration, we should have obtained $$x \simeq \frac{6 \log \left(\frac{9}{8}\right) \log (6)-3 \log (384)}{3 \log ^2\left(\frac{3}{2}\right)-\log (147456)}$$

For the second root, using $x_0=4.0$, the successive iterates are $3.91886$, $3.91689$ which is the solution for six significant figures. Limiting to a single iteration, we should have obtained $$x \simeq \frac{\log \left(\frac{4096}{81}\right)+\log (6) \log (64)}{\log ^2(4)+\log \left(\frac{8}{3}\right)}$$ As you can see, all of this could be done using a pocket calculator.