Write: 2^{2x}+6\cdot 2^x +18=(2^x+3)^2+9 And see that the minimum happens when 2^x \rightarrow 0 once 2^x >0. So in this case we have infimum (not minimum) and it is 18.

\displaystyle{x}={{\log}_{{5}}{\left({2}\right)}} Explanation: \displaystyle{\left({5}^{{2}}\right)}^{{x}}-{8}\cdot{5}^{{x}}=-{16} \displaystyle{t}^{{2}}-{8}{t}=-{16} \displaystyle{t}={4} ...

Assuming you mean 2^{2x} - 3\times2^x - 10 = 0, then since this is a quadratic in 2^x we can factorise as (2^x+2)(2^x-5)=0. Since 2^x>0 \forall x, 2^x+2\neq0, so we must have 2^x=5. ...

You can still convert it into a polynomial, since 2^{5x} - 8 = 2\cdot 2^{2x} converts to y^5-8=2y^2 if you introduce y=2^x .

When you have ab \gt 0, you can have a\gt 0, b \gt 0 or a \lt 0, b \lt 0. So for your first example you should have [3^x \gt 9 and 3^x \gt 1] or [3^x \lt 9 and 3^x \lt 1]. You can ...

Your notes apparently have an error. The series \sum_{n=1}^{\infty} \frac{(-1)^{n-1} (n-1)!}{(1+x)^n} is first of all not a Taylor series, and second of all, converges nowhere by the ratio test: ...