answer.md

$$ We are given $( f(x) = \log(3e^{2x} + 4x - 5) )$, and we need to determine the behavior of ( f(x) ) as $( x \to +\infty )$.

Step 1: Simplify $( 3e^{2x} + 4x - 5 ) as ( x \to +\infty )$

For very large $( x )$, the term $( 3e^{2x} )$ dominates over $( 4x - 5 )$, because $( e^{2x} )$ grows much faster than any polynomial. Hence: $$[ 3e^{2x} + 4x - 5 \sim 3e^{2x} \quad \text{as } x \to +\infty. ]$$

Step 2: Take the logarithm

Using the logarithmic property $( \log(ab) = \log(a) + \log(b) ), we can write: $$[ \log(3e^{2x} + 4x - 5) = \log(3e^{2x}) + \log\left(1 + \frac{4x - 5}{3e^{2x}}\right). ]$$

Simplify each term:

$( \log(3e^{2x}) = \log(3) + \log(e^{2x}) = \log(3) + 2x )$.
For the second term $( \log\left(1 + \frac{4x - 5}{3e^{2x}}\right) )$:
- As $( x \to +\infty ), ( \frac{4x - 5}{3e^{2x}} \to 0 ), so we can approximate ( \log(1 + u) \sim u ) for small ( u ).
- Thus, ( \log\left(1 + \frac{4x - 5}{3e^{2x}}\right) \sim \frac{4x - 5}{3e^{2x}} \to 0 ) as ( x \to +\infty ).

Step 3: Combine terms

The dominant term is: [ f(x) = 2x + \log(3) + o(1), \quad \text{where } o(1) \text{ denotes a term that tends to 0 as } x \to +\infty. ]

Step 4: Interpret the options

From the analysis:

( f(x) = 2x + \log(3) + o(1) ) as ( x \to +\infty ). This matches option (a).
The asymptote is ( y = 2x + \log(3) ), so option (d) is correct regarding the asymptote.

Final Answer:

The correct answers are: [ \boxed{\text{(a) and (d)}} ]