It is possible to solve analytically this kind of equation: $$ a\cdot x^{\frac{4}{5}} - bx = c $$ for $x$ ?
I thought about applying for both side logarithm with base equal $\frac{4}{5}$.
But what the next step should be?
Or maybe it is not possible to solve this kind of equation by hand?
I will be thankful for any clue and help.
On
No, these don't allways have analytical solution, take a look at this https://en.m.wikipedia.org/wiki/Galois_theory, set $ y = x^{1}{5} $ then this equation becomes $ a\cdot y^4 - b\cdot y^5 - c = 0 $, which is a quintic function !
On
Your equation is equivalent to a quintic of the form $x^5 + Ax + B = 0$, which is known as the Bring–Jerrard form. Analytical solutions to such equations may sometimes be found, depending on the coefficients. See the Wikipedia link for some information on how to determine if a particular equation can be solved that way.
However, if you're willing to use non-elementary functions, quintics can be solved analytically via Bring radicals, which are just the real roots of $t^5 + t - a=0$ for real $a$. And they can be solved using elliptic functions. Please see the Wikipedia article (and its links) for further details.
$$(1) \quad a \cdot x^{4/5}-b \cdot x=c$$
If you've never heard of U-substitution. It's time to learn, it's very useful. Choose $u=x^{1/5}$ and substitute into $(1)$.
$$(2) \quad a \cdot (u^5)^{4/5}-b \cdot u^5=c$$
$$\Rightarrow u^4 \cdot (a-b \cdot u)=c$$
This could be solved, but it'd be very tedious and depend on the constants. I suggest newton's method, if you know calculus, or fixed point iteration for numerical values. Using the later we have,
$$u_{n+1}=\left(\cfrac{c}{a-b \cdot u_n} \right)^{1/4}$$
Which may converge for suitable choice of $u_0$ to one of the roots. You'd need to know a lot more about the subject to prove it converges. However, for $a=2$ $\ b=5$ and $c=7$ we have one of roots, of the original equation, given by,
$$\lambda=\cfrac{7^{5/4}}{\left(2-5 \cdot \left(\cfrac{7}{2-5 \cdot \left(\cfrac{7}{2-5 \cdot \left(\cfrac{7}{2-5 \ddots} \right)^{1/4}} \right)^{1/4}} \right)^{1/4} \right)^{5/4}}$$