Logarithmic Question?

OK, here you go sweetheart :)


f(x) = log₄ (x + 1) + log₄ (4x + 3)


We use 'log laws' to simplify this :

f(x) = log₄ [(x + 1)(4x + 3)]

f(x) = log[(x + 1)(4x + 3)] /log4


We now set this equal to zero and solve for 'x' :

f(x) = 0
log[(x + 1)(4x + 3)] /log4 = 0
log[(x + 1)(4x + 3)] = 0
[(x + 1)(4x + 3)] = 10⁰
(x + 1)(4x + 3) = 1
4x² + 7x + 3 = 1
4x² + 7x + 2 = 0


We now use the quadratic formula as follows :


x = [- 7 ±√(7² - 4×4×2)] /(2×4)
x = (- 7 ±√17) /8


Hence the two solutions are :

x₁ = (- 7 - √17)/8 ≈ - 1.390388
x₂ = (- 7 + √17)/8 ≈ - 0.359612


We now substitute BOTH solutions to check if they still make sense :


When x ≈ - 1.390388, f(x) is undefined as only positive values can be taken for the logarithmic function to be defined.


Hence, it stands to reason, there is ONLY one real solution.


That solution is :

x = (- 7 + √17)/8 ≈ - 0.359612


Hope this helps !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

log₄(x + 1) + log₄(4x + 3) = 0
log₄[(x + 1)(4x + 3)] = log₄(1)
(x + 1)(4x + 3) = 1
4x² + 7x + 3 = 1
4x² + 7x + 2 = 0
x = [-7 ± √(7² - 4*4*2)]/(2*4)
x = (-7 + √17)/8 or x = (-7 + √17)/8 (rejected for x + 1 < 0)

Hence, x = (-7 + √17)/8

log₄(x+1) + log₄(4x+3) = 0
log₄((x+1)(4x+3)) = 0
log₄(4x² + 7x + 3) = 0

4x² + 7x + 3 = 1
4x² + 7x + 2 = 0

quadratic formula
x = [-7 ± √(7² – 4·4·2)] / [2·4]
 = [-7 ± √17] / 8
 ≅ -0.359, -1.39
-1.39 is an extraneous solution
x = -0.359

log4(x+1)+log4(4x+3) = 0
log4 ((x+1)√(4x+3)) = 0
(x+1)(4x+3) = 4^0
4x^2+7x+3 = 1
4x^2+7x+2 = 0

This equation is of form ax^2+bx+c=0
a = 4 b = 7 c = 2
x=[-b+/-sqrt(b^2-4ac)]/2a]
x=[-7 +/-sqrt(7^2-4(4)(2)]/(2)(4)
discriminant is b^2-4ac =17
x=[-7 +√(17)] / (2)(4)
x=[-7 -√(17)] / (2)(4)

x = (-7+√17)/8 --->-0.36
x = (-7-√17)/8 --->-1.39

x = -1.39 would produce log of a negative number so it can't be.

x = (-7+√17)/8