Q:

Solving an inequality: $\;2^{x^2} \cdot 3^x < 6$

Can you show me please how can I solve the following exercise:
$$2^{\large x^2} \cdot 3^x < 6$$
Thank you , in advance :) 

A:

Take the logarithm of each side of the inequality, and recall the laws of exponents:  $$\log(a^b) = b\log(a), \\ \\ \log(a \cdot b) = \log a + \log b$$

$$\begin{align} 2^{\large x^2} \cdot 3^x < 6 & \iff \log(2^{x^2} \cdot 3^x) < \log 6 \\ \\ &\iff \log (2^{x^2}) + \log (3^x) < \log 6 \\ \\ & \iff x^2 \log(2) + x\log(3) < \log 6\\  \\& \iff x^2 \log(2) + x\log(3) - \log 6 \lt 0 \\ \\ &\iff x^2 \log(2) + x\log(3) - (\log(2\cdot 3))< 0 \\ \\ &\iff x^2 \log(2) + x\log(3) - (\log 2 + \log 3) \lt 0\end{align}$$
Clearly, you have a quadratic in $x$. Solve for when the left-hand side equals the right-hand side to find the endpoints of intervals to test, and determine from there the interval(s) on $x$ which satisfies the given inequality.

