How can i solve $5^{2x-\frac{1}{3}x^2} < 5^{2-2x} * (5^\frac{1}{3})^{x^2}+24$? How can i solve it?
$$5^{2x-\frac{1}{3}x^2} < 5^{2-2x} * (5^\frac{1}{3})^{x^2}+24$$  
I don't have idea how to solve it..

We have,
$$5^{2x-\frac{1}{3}x^2} < 5^{2-2x} * (5^\frac{1}{3})^{x^2}+24$$  
$$5^{2x-\frac{1}{3}x^2} - 5^{2-2x} \cdot (5^\frac{1}{3})^{x^2}- 24<0 $$  
$$5^{2x-\frac{1}{3}x^2} - 5^{2-2x + \frac{1}{3} x^2 } - 24 < 0 $$  
Substitute $ 5^{2x - \frac{1}{3}x^2} = t $,
$$t - \frac{5^2}{t} - 24 < 0 $$  
$$\frac{(t-25)(t+1)}{t} < 0 $$
$$ t \in (-∞,-1) \cup (0,25) $$
Since $ t$ is a constant raised to some exponent , it can't be negative.
So, $$ t \in (0,25) $$
Which implies,
$$5^{2x-\frac{1}{3}x^2} < 25 $$
To get $x$ taking logarithm with base as $5$,
$$\log_5{5^{2x-\frac{1}{3}x^2} } < \log_5 {5^2} $$
$$ 2x- \frac{1}{3}x^2 < 2 $$
$$ - \frac{1}{3}x^2 + 2x -3 < 0  $$
$$ \frac{1}{3}x^2 -2x + 3 > 0  $$
$$ x^2- 6x + 6 > 0  $$
$$ x = 3+√3, x= 3-√3 $$
