# Algebra exponential problem with 2 different bases

$\sqrt[{3x-1}]{\sqrt[{8-3x}]{(-x)^x}}=\sqrt[{5x}]{2x}$

Any ideas how could i approach this?

I got it to this point:

$(-x)^\frac x{(3x-1)(8-3x)}=(2x)^\frac1{5x}$

Thanks for the answer (x = 2) but if we graph this function why we only have one point intersect that is in in x = 0 and the f(x) is equal to 1?

What's going on here?

If $x$ is negative, the RHS does not make sense $($'minus one-th root'$)$. If $x$ is positive, then the LHS does not make sense unless:
$$3x-1 > 0 \iff x>\frac13$$ $$8-3x>0\iff x<\frac83$$ $$x \text{ is an integer}$$
where the last condition is necessary in order for $(-x)^x$ to make sense. This would leave us with $x=1,2$ as our only options.
We can easily check that $x=1$ is no good, but $x=2$ does the trick.
• Graphing this function does not make much sense because, for most values of $x$, it is not well defined. – Fimpellizieri Mar 1 '18 at 17:20