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..
 A: Taking uldek's comment in the form of an answer.
Let $A=5^{2x-{x^2}/3}$. Note that $A>0$ for any $x$.

Notice that A appears as is on the LHS.
Notice that A is hidden on the RHS, as $5^{2-2x}*(5^{1/3})^{x^2}=5^2*5^{-2x+{x^2}/3}=5^2*A^{-1}$.

Now the inequality is $A < 25*A^{-1} +24$;

Take all the terms on the same side:  

$A -24 - 25*A^{-1} < 0$

As $A>0$, you can multiply both sides by A without changing the sense of the inequality:  

$A^2 - 24A - 25 < 0$

Factor:  

$(A-25)(A+1)<0$  

Draw a little table where you show the sign of the product:  

$$
\begin{array}{c|ccc}
A & \text{under -1} & \text{between} & \text{over 25} \\
\hline
A+1 & - & + & + \\
A-25 & - & - & + \\
(A-25)(A+1) & + & - & + \\
\end{array}
$$

So, that the inequation is negative implies $-1<A<25$
Remember that $A>0$ no matter what, so we solve:  

$0<A<25$
$5^{2x-{x^2}/3}<25$

We notice that $25=5^2$

$5^{2x-{x^2}/3}<5^2$  

By definition of what $5^r$ is (that is : $5^r=e^{r*ln(5)}$), and the exponential function being a strictly increasing function, so $e^X<e^Y$ is equivalent to $X<Y$.
  Put another way, "you take the logarithm" of the inequality - logarithm being also a strictly increasing function.

$2x-{x^2}/3<2$  

Put all the terms on the same side of the inequality (this time, I am choosing the right hand side):

${x^2}/3-2x+2>0$  

Multiplying by 3 > 0 keeps the sense of the inequality:  

${x^2}-6x+6>0$  

Factor... yes, discuss $b^2-4ac$, that stuff...

$(x-(3+\sqrt{3}))(x-(3-\sqrt{3}))>0$

Discuss the sign of the product...

$$
\begin{array}{c|ccc}
x & \text{under $3-\sqrt{3}$} & \text{between} & \text{over $3+\sqrt{3}$} \\
\hline
x-(3-\sqrt{3}) & - & + & + \\
x-(3+\sqrt{3}) & - & - & + \\
(x-(3+\sqrt{3}))(x-(3-\sqrt{3})) & + & - & + \\
\end{array}
$$

You can now write down the conclusion.

Hope this helps!
(and I hope that no typo sneaked in my text)
A: 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 $5$,
$$\log_5{5^{2x-\frac{1}{3}x^2} } < \log_5 {5^2} $$
$$ 2x- \frac{1}{3}x^2 < 2 $$
$$ - \frac{1}{3}x^2 + 2x -3 < 0  $$
$$ \frac{1}{3}x^2 -2x + 3 > 0  $$
$$ x^2- 6x + 6 > 0  $$
$$ x = 3+√3, x= 3-√3 $$
