ln(e^(5)·3^log3(e)+1/2e^-3ln(2)

Question

the first base e has en exponent of 5 times 3 to the power/exponent of log with a base of 3 (3). the second base of e has a power/exponent of (-3In(2)

Answer 1

recall that
log_a(b) = 1/log_b(a)

so, your expression is

ln(e^5) + ln(e) + (1/2)(1/8))
= 5+1+1/16
= 97/16