$\log_{49}({x^2})=\log_{2}({5x-8})$ I need help with this problem.
$$\log_{49}({x^2})=\log_{2}({5x-8})$$
I've tried doing this:
$$\log_{7}({x})=\log_{2}({5x-8})$$
$$\frac{\log x}{\log7}=\frac{\log (5x-8)}{\log2}$$
$$\log2 \log x=\log(5x-8)\log7$$
$$\frac{\log(5x-8)}{\log x}=\frac{\log2}{\log7}$$
$$\log_x (5x-8)=\frac{\log2}{\log7}$$
$$5x-8=x^{\frac{\log2}{\log7}}$$
I'm stuck here. I did a Newton-Raphson and got $1.8489491136680562534$. I need a closed form solution, though.
 A: Possibly the original problem was mistyped. If the equation
$\log_4(\color{blue}{9}x^2)=\log_2(5x-8)$
was meant, that has an easy solution ... if you watch your signs and remember that the argument of a logarithm must be positive (if you are working with real variables).

With the proposed alternative, $\log_2(\pm(3x))=\log_2(5x-8)$, thus $\pm3x=5x-8$ and $x=1$ or $x=4$; but only $x=4$ gives positive arguments for both original logarithms.

A: Since you copied correctly, there is no closed form solution for the zero of function$$f(x)=5x-8-x^{\frac{\log(2)}{\log(7)}}$$ I suppose that, graphing, you noticed that the solution is just smaller than $x=2$.
Let $a=\frac{\log(2)}{\log(7)}$ and expand around $x=2$
$$f(x)=\left(2-2^a\right)+\left(5-2^{a-1} a\right) (x-2)-2^{a-3} (a-1) a
   (x-2)^2-\frac{1}{3} \left(2^{a-4} (a-2) (a-1) a\right) (x-2)^3+O\left((x-2)^3\right)$$ Using series reversion
$$x=2+\frac{2^a-2+f(x)}{5-2^{a-1} a}-\frac{2^{a-2} (a-1) a
   \left(2^a-2+f(x)\right)^2}{\left(5-2^{a-1} a\right)^2 \left(2^a
   a-10\right)}+O\left(\left(2^a-2+f(x)\right)^3\right)$$ which is the same as the first iteration of Halley method.
Make $f(x)=0$ as desired and $a=\frac{\log(2)}{\log(7)}$; so, you have an explicit formula and, converted to decimals
$$x_{(3)}=\color{blue}{1.8489}5718257\cdots$$ Do the same with Householder method to obtain
$$x_{(4)}=\color{blue}{1.848949}5426\cdots$$ while, as you wrote, the solution is
$$x_{(\infty)}=\color{blue}{1.8489491137}\cdots$$
Edit (just for the fun)
$$f(x) \sim 5x -8 -x^{\frac 1 3}$$ the solution of which giving
$$x_0=\frac{8}{15 \sqrt{15}}\Bigg[\cosh \left(\frac{1}{3} \cosh ^{-1}\left(12 \sqrt{15}\right)\right) \Bigg]^3=1.84531$$
which is a much better starting point for any Newton-type method or approximation.
Using the implicit function theorem
$$x_1=x_0+\frac{x_0^{a_0}\, \log (x_0)}{5-a_0\, x_0^{a_0-1}} (a-a_0)$$
Numerically, $x_1=1.84891$
