Explanation:
logarithm rules :
logb(N×M) = logb(N) + logb(M).
that goes both ways, of course.
b^logb(k) = k
logb(M^k) = k×logb(M)
this goes also both ways, of course.
ln = loge (base e)
(i)
ln(x) + ln(7x + 1) = 1
ln(x(7x + 1)) = 1
ln(7x² + x) = 1
so, we are putting everything to the power of e :
e^ln(7x² + x) = e¹
7x² + x = e
7x² + x - e = 0
a quadratic equation
ax² + bx + c = 0
has the general solution
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 7
b = 1
c = -e
x = (-1 ± sqrt(1² - 4×7×-e))/(2×7) =
= (-1 ± sqrt(1 + 28e))/14
x1 = (-1 + sqrt(1 + 28e))/14
x2 = (-1 - sqrt(1 + 28e))/14
since x2 is negative, it is disqualified, as the log of a negative number is undefined.
so, x1 is our only solution.
(ii)
2ln(x + 1) - ln(4x) = 0
2ln(x + 1) = ln(4x)
ln((x + 1)²) = ln(4x)
putting everything to the power of e :
(x + 1)² = 4x
x² + 2x + 1 = 4x
x² -2x + 1 = 0
to solve the quadratic equation as in (i) :
a = 1
b = -2
c = 1
x = (2 ± sqrt((-2)² - 4×1×1))/(2×1) =
= (2 ± sqrt(4 - 4))/2 = (2 ± 0)/2 = 2/2 = 1
x = 1 is our only solution.