# ln e^(x+2) = 5 + ln e^2

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ln e^(x+2) = 5 + ln e^2

We know that:

ln e^a = aln e = a*1 = a

==> ln e^(x+2) = 5 + ln e^2

==> (x+2) = 5 + 2

==> x+2 = 7

==> x= 7-2

==> x= 5

The equation ln e^(x+2) = 5 + ln e^2 has to be solved.

ln e^(x+2) = 5 + ln e^2

ln e^(x+2) - ln e^2 = 5

Use the property of logarithm log a - log b = log(a/b)

`ln(e^(x+2)/e^2) = 5`

The logarithm y of a number x to base b is defined as `log_b x = y` , if `x = b^y`

As ln is logarithm to base e, `ln(e^(x+2)/e^2) = 5` gives:

`e^(x+2)/e^2 = e^5`

`e^(x+2 - 2) = e^5`

Equate the exponent as the base is the same.

This gives the solution of the equation x = 5.

For the beginning, we'll use the power property of logarithms:

ln e^(x+2) = (x+2)*lne, where ln e = 1

ln e^(x+2) = x+2

ln e^2 = 2*ln e = 2

Now, we'll rewrite the equation:

x+2 = 5+2

We'll eliminate like terms:

**x = 5**

Another manner of solving would be to subtract ln e^(x+2) both sides:

0 = 5 + ln e^2 - ln e^(x+2)

We'll subtract 5 both sides:

ln e^2 - ln e^(x+2) = -5

We'll use the quotient property:

ln [e^2/e^(x+2)] = -5

But e^2/e^(x+2) = e^(2-x-2) = e^-x

ln e^-x = -5

-x*ln e = -5

-x = -5

**x = 5**

lne^(x+2) = 5+lne^2.

To solve for x.

Solution:

We eearrange the given equation as:

ln e^(x+2) - lne^2 = 5.

ln {[e^(x+2)]/e^2} = 5

ln{ e ^(x+2-2)} = 5 = e^5, as 5 = lne^5 by definition of logarithms.

ln e^x = lne^5

e^x = e^5 by one to one property

x =5 by one tone property.