1) Factor the equation first:
\(2e^{2x} + 4xe^{2x} = 0\)
\(e^{2x}(2 + 4x) = 0\)
So either \(e^{2x} = 0\) or \(2 + 4x = 0\). But \(e^{2x} = 0\) can never equal 0! So, we only need to solve for the second equation:
\(2 + 4x = 0\)
\(\bf x = {-\frac{1}{2}}\)
2) First, single out the term with the exponent:
\(2 + 3(4^{2x - 1}) = 43\)
\(3(4^{2x - 1}) = 41\)
\(4^{2x - 1} = \frac{41}{3}\)
Now, taking the log base 4 of each side gives us
\(2x - 1 = \log_4 \frac{41}{3}\)
and we can now solve for x:
\(2x - 1 = \log_4 \frac{41}{3}\)
\(2x = 1 + \log_4 \frac{41}{3}\)
\(2x = \log_4 4 + \log_4 \frac{41}{3}\)
\(2x = \log_4 \frac{164}{3}\)
\(x = \frac{1}{2} \log_4 \frac{164}{3}\)
\(x = \log_4 \sqrt \frac{164}{3}\)
\(\bf x = \log_4 2 \sqrt \frac{41}{3}\)
