3

I have been playing with graphs until made a nice equation

$$\log_\sqrt[34]{2} x = 4x^4-3x^3-2x^2+x$$


The real answers are 1 and 2. But how to solve it? And is it possible to stretch it to complex values?
I'm relatively new to math as such, so would appreciate as much explanations as possible.
I'm also pretty sure that you'd need to use the product log to solve it. I tried to simplify it and got
$$34\frac{\ln x}{\ln 2} = x(x-1)(4x^2+x-1)$$
Sebastiano
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A PIG
  • 39

2 Answers2

3

You can at least have a very good approximation of it.

Multiplied by $x$, the first derivative of $\text{(rhs - lhs)}$ is $$16 x^4-9 x^3-4 x^2+x-\frac{34}{\log(2)}$$ which can be solved using radicals. Converted to decimals, its positive zero is $$x_*=1.539220361840668443650\cdots$$ The second derivative test shows that it corresponds to a maximum.

Since we already needed to solve a quartic equation, expanding $$\text{(rhs - lhs)}=f(x)=x(x-1)(4x^2+x-1)-\frac{34}{\log (2)}\, \log(x)$$ around $x_*$ to fourth order, the two real approximate zero's are $$x_1=0.998345 \qquad \text{and} \qquad x_2=1.999688$$

Claude Leibovici
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1

Graph of a concave up function intersects the graph of a concave down function on a closed interval at at most 2 points.

Let $f(x)=\frac{36}{\ln 2}\ln x$ and $g(x)=4x^4-3x^3-2x^2+x$. Then, $f''(x)=-\frac{36}{x^2\ln2}<0$ and so $f(x)$ is concave down on its domain which is $x>0$.

On the other hand, $g''(x)=48x^2-18x-4.$ The roots of $g''(x)$ are $\frac{9\pm{\sqrt{273}}}{48}.$ So, g(x) is concave up for $x\geq\frac{9+\sqrt{273}}{48}\approx0.53172$ and we conclude that there are two solutions of $f(x)=g(x)$ on this interval.

On $(0,\frac{9+\sqrt{273}}{48})$ both are concave down. We need another analysis here. Or by a careful curve sketching we can see that the graph of the quartic stays above of the logarithm. No solition here.

Bob Dobbs
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