Logarithmic inequality with variable base

Question

If $$\log_{3x+5} (ax^2+8x+3)>2$$ find the interval in which x lies.

Answer is $\left(-\dfrac43 , -\dfrac{23}{22}\right)$. Now hint me how to solve the problem please.

Answer 1

In general, if $b > 1$ then $$\log_b A > k \iff b^k < A$$ while if $0 < b < 1$ then $$\log_b A > k \iff b^k > A$$

Applying this to your problem, we have two cases:

$$3x+5 > 1 \textrm{ and } (3x+5)^2 < ax^2 + 8x +3$$ or $$0 < 3x+5 < 1 \textrm{ and } (3x+5)^2 > ax^2 + 8x +3$$

Can you take it from there?

Answer 2

HINT

Recall that $$\log_b a = \frac{\ln a}{\ln b} \quad \text{and} \quad c\ln a=\ln \left(a^c\right),$$ so can you plug this in, simplify your inequality and express it as a condition on both polynomials in $x$?