Related Formula
ₐ(bc) = ₐ b + ₐ c$$\log_a(bc) = \log_a b + \log_a c$$
b(aⁿ) = n b a$$\log_b(a^n) = n\log_b a$$
ₐ b = (1)/( b a)$$\log_a b = \frac{1}{\log_b a}$$
Core Logic
Factor the arguments in the logarithmic equation:
6x² + 28x + 30 = (x+3)(6x+10)$6x^2 + 28x + 30 = (x+3)(6x+10)$
x² + 6x + 9 = (x+3)²$x^2 + 6x + 9 = (x+3)^2$
Substitute these into the equation:
(x+3)[(x+3)(6x+10)] = 5 - 2 (6x+10)(x+3)²$$\log_{(x+3)}[(x+3)(6x+10)] = 5 - 2\log_{(6x+10)}(x+3)^2$$
1 + (x+3)(6x+10) = 5 - 4 (6x+10)(x+3)$$1 + \log_{(x+3)}(6x+10) = 5 - 4\log_{(6x+10)}(x+3)$$
Step 1: Variable Substitution
Let A = (x+3)(6x+10)$A = \log_{(x+3)}(6x+10)$. The equation transforms to:
1 + A = 5 - (4)/(A)$$1 + A = 5 - \frac{4}{A}$$
A + (4)/(A) = 4$$A + \frac{4}{A} = 4$$
A² - 4A + 4 = 0 (A - 2)² = 0 A = 2$$A^2 - 4A + 4 = 0 \implies (A - 2)^2 = 0 \implies A = 2$$
Substitute A = 2$A = 2$ back:
(x+3)(6x+10) = 2$$\log_{(x+3)}(6x+10) = 2$$
6x + 10 = (x+3)²$$6x + 10 = (x+3)^2$$
6x + 10 = x² + 6x + 9$$6x + 10 = x^2 + 6x + 9$$
x² = 1 x = ± 1$$x^2 = 1 \implies x = \pm 1$$
Step 2: Checking Domain Validity
For x = 1$x = 1$:
Base x+3 = 4 > 0, ≠ 1$x+3 = 4 > 0, \neq 1$.
Base 6x+10 = 16 > 0, ≠ 1$6x+10 = 16 > 0, \neq 1$.
Valid solution.
For x = -1$x = -1$:
Base x+3 = 2 > 0, ≠ 1$x+3 = 2 > 0, \neq 1$.
Base 6x+10 = 4 > 0, ≠ 1$6x+10 = 4 > 0, \neq 1$.
Valid solution.
Sum of all real roots = 1 + (-1) = 0$= 1 + (-1) = 0$.
Pattern Recognition
When dealing with logarithms containing polynomial bases and arguments, always check if they are directly factorable into each other. A substitution like A + B/A = C$A + B/A = C$ will often emerge.
Chapter Mix
Class 11 Maths: Quadratic Equations
Class 11 Maths: Functions
