# Is -2ln e^2 + 1 = -3 ?

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We need to find the value of -2*ln e^2 + 1.

-2*ln e^2 + 1

Use the relation log b^a = a* log b

=> (-2*2)*ln e + 1

ln indicates the logarithm is to the base e, and the log of the base is equal to 1. So ln e = 1

=> -4*1 + 1

=> 1 - 4

=> -3

**The value of -2*ln e^2 + 1 is indeed -3**

Let's verify the given relation:

First, we'll use the power property of logarithms:

-2ln e^2 = ln (e^2)^-2 = ln 1/e^4

We'll write 1 = ln e

We'll re-write the relation:

ln 1/e^4 + ln e = -3

We'll use the product property of logarithms:

ln (1/e^4)*e = -3

ln e/e^4 = -3

ln 1/e^3 = -3

ln e^-3 = -3

-3ln 3 = -3

-3 = -3

**So, the given identity is true: -2ln e^2 + 1 = -3.**