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Need help with this question-example-1
User Lucasgcb
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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.

User Criz
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