# simplify Log3e(2e)the 3e is the base i guess

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### 1 Answer

Since the problem specifies that `3e` represents the base of the logarithm, you may use the following logarithmic identities, such that:

`log(a*b) = log a + log b`

`log_a b = 1/(log_b a)`

Reasoning by analogy yields:

`log_(3e) (2e) = log_(3e) 2 + log_(3e) e`

Using the identity `log_a b = 1/(log_b a)` for the term` log_(3e) e` yields:

`log_(3e) (2e) = log_(3e) 2 + 1/(log_e (3e))`

You need to remember that you may conventionally write `(log_e (3e)) ` as `ln(3e)` .

Using the identity `log(a*b) = log a + log b` for `ln(3e)` yields:

`ln(3e) = ln 3 + ln e`

You need to substitute 1 for `ln e` such that:

`ln(3e) = ln 3 + 1`

`log_(3e) (2e) = log_(3e) 2 + 1/(ln 3 + 1)`

Using the identity `log_a b = 1/(log_b a)` for the term `log_(3e) 2 ` yields:

`log_(3e) 2= 1/(log_2 (3e))`

`log_(3e) 2= 1/(log_2 3 + log_2 e)`

`log_(3e) 2= 1/(log_2 3 + 1/(ln 2))`

`log_(3e) (2e) = 1/(log_2 3 + 1/(ln 2)) + 1/(ln 3 + 1)`

**Hence, using the logarithmic identities, you may evalaute the given logarithm such that **`log_(3e) (2e) = 1/(log_2 3 + 1/(ln 2)) + 1/(ln 3 + 1).`