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I have tried separating using exponential rules couldn't solve that way. Next I tried using log to bring down the exponent, still couldn't do it.

$$5x^{7x+3}>5^{-3}$$

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    It seems to be false. $x^{7x+3}$ is zero at the limit $x\rightarrow 0$. – 2017-02-24

2 Answers 2

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This doesn't seem like being analytically solvable. Nevertheless, if you manage to show for example that the left-hand-side function is strictly increasing, you could be done be guessing the equality case.

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You can change to base "e" rather than x using $a^b= e^{b ln(a)}$ so you can write $5x^{7x+ 3}= 5e^{(7x+ 3)ln(x)}$. Since a region satisfying an inequality always has the region where the equality is satisfied, would first look at $5x^{7x+3}= 5e^{(7x+ 3)ln(x)}= 5^{-3}$. Of course, then, $e^{(7x+3)ln(x)}= 5^{-4}= \frac{1}{625}$. Taking the natural logarithm of both sides of that $(7x+ 3)ln(x)= ln\left(\frac{1}{625}\right)= -6.43775$. As Tig La Pomme said, that can not be solve "analytically" (in terms of the usual functions though you might be able to get it into a form you could solve using the "Lambert W function" which is defined as "the inverse function to $f(x)= xe^x$.