d) To solve exponential equations or logarithmic equations, we
have some commonly used methods:
i) apply the equivalent relation $y = a^x \iff x = \log_a y$
ii) apply properties so that we can have the form of exponent
equals exponent (logarithm = logarithm) with the same base so
that we can simplify the equation to linear or quadratic.
iii) use substitution to change the original equation into
quadratic equation
Now use the two methods above to solve the following equations:
$2^{5x-1} = 3$
$\log_6 x + \log_6 (x - 1) = 1$
$e^{2x} - 3e^x = 10$
