# Logarithmic inequality with variable base [closed]

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.

## closed as off-topic by Saad, Jyrki Lahtonen, user416281, Christopher, Trần Thúc Minh TríJun 22 '18 at 13:22

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• How can the solution be independent of $a$? – Yanko Jun 22 '18 at 11:57

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?

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 $x$?

• But how do i deal with that a in the equation? I have simplified it to an extent but there is an a in the equation. Now what do I do? – Wonder Jun 22 '18 at 4:56