Let f(x)=ln(x+2e)+ln(x), x>0 Can anyone teach me how to solve this M1 question? Thx?

　
Questions：
1.
Let f(x) = ln(x + 2e) + ln(x), x > 0
Q1) find f(2e)
Q2) solve f(x) = 1

2.
Solve e²ˣ⁺¹ - 4eˣ⁺¹ - 5e = 0
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Solutions：
1Q1)
Sub x = 2e into f(x) = ln(x + 2e) + ln(x)
f(2e)
= ln(2e + 2e) + ln(2e)
= ln(4e) + ln(2e)
= [ln(4) + ln(e)] + [ln(2) + ln(e)]
= ln(2²) + 1 + ln(2) + 1
= 2ln(2) + ln(2) + 2
= 3ln(2) + 2

1Q2)
ln(x + 2e) + ln(x) = 1
ln[(x + 2e)x] = ln(e)
x² - 2ex = e
x² - 2ex - e = 0
x = {-(2e) ± √[(2e)² - 4(-e)]}/2
x = [-2e ± 2√(e² + e)]/2
x = -e + √(e² + e) or -e - √(e² + e) ( rejected )

2.
e²ˣ⁺¹ - 4eˣ⁺¹ - 5e = 0
e²ˣe - 4eˣe - 5e = 0
e²ˣ - 4eˣ - 5 = 0
(eˣ - 5)(eˣ + 1) = 0
eˣ = 5 or eˣ = -1 ( rejected )
eˣ = 5
ln(eˣ) = ln(5)
x = ln(5)
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Remarks：
A)
Properties of NATURAL LOGARITHM ( ln )：
For a, b > 0,
ln(1) = 0
ln(e) = 1
ln(ab) = ln(a) + ln(b)
ln(aˣ) = x ln(a)

B)
For a quadratic equation,
ax² + bx + c = 0 ...... where a ≠ 0
x = [-b ± √(b² - 4ac)]/(2a)