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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..

Master Question
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    See https://math.stackexchange.com/questions/2570576/how-can-i-solve-this-42x-15-4x2-16x2-fracx2-0 – lab bhattacharjee Dec 17 '17 at 15:59
  • I asked it, but i don't understand how to solve it according to the older exercise. – Master Question Dec 17 '17 at 16:01
  • Rearrange to form a quadratic equation like the linked answer. Now $$(x-a)(x-b)<0\implies a<x<b$$ – lab bhattacharjee Dec 17 '17 at 16:02
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    Put $A=5^{2x-x^2/3}$. Then the inequality is the same as $A<5^2/A+24$, or $\frac{(A-25)(A+1)}{A}<0$. You get that $A=5^{2x-x^2/3}$ must be either on $(-\infty,-1)$ or in $(0,25)$. $A$ cannot be negative, therefore it is in $(0,25)$. Take logarithms base $5$ to conclude that $2x-x^2/3<2$. Factor, and solve. –  Dec 17 '17 at 16:06
  • @ uldek, you should post this as an answer, not a comment. Then we can up vote your solution. – nonuser Dec 17 '17 at 16:10
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    @uldek The fact remains, that kind of statement belongs in an answer, not in a comment. You have a habit of answering questions in comments, I've noticed. See the guidelines for commenting if you need any further clarification on the matter. – Robert Howard Dec 17 '17 at 16:55
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    " I am not interested in your vote. It means nothing to me." And we are not interested in your desires and wishes. They mean nothing to us. We want the questioned answered. – fleablood Dec 17 '17 at 17:29

2 Answers2

2

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)

Taamer
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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 $5$,

$$\log_5{5^{2x-\frac{1}{3}x^2} } < \log_5 {5^2} $$

$$ 2x- \frac{1}{3}x^2 < 2 $$

$$ - \frac{1}{3}x^2 + 2x -3 < 0 $$

$$ \frac{1}{3}x^2 -2x + 3 > 0 $$

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

$$ x = 3+√3, x= 3-√3 $$

Fghj
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