ln e^(x+2) = 5 + ln e^2
4 Answers
hala718 | High School Teacher | (Level 1) Educator Emeritus
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
tonys538 | Student, Undergraduate | (Level 1) Valedictorian
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.
giorgiana1976 | College Teacher | (Level 3) Valedictorian
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
neela | High School Teacher | (Level 3) Valedictorian
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.