Find all real values of $x$ satisfying the inequality: $x^{\log_3 (x+4)} \lt 3^5$.
2026-04-28 15:25:05.1777389905
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Find solutions of the inequality
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We need to solve $f(x)<5$, where $f(x)=\log_3x\log_3(x+4)$.
But $f$ is an increasing function on $[1,+\infty)$ and all $0<x<1$ are solutions.
Thus, the answer is $(0,x_0)$, where $x_0$ is a root of the equation $f(x)=5$:
$x_0=9.9024...$
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Take base 3 logarithms on both sides. Sign does not change as function is increasing:- $$(\log_3x)(\log_3(x+4))<5$$
The function $(\log_3x)(\log_3(x+4))$ is increasing. Solving $(\log_3x)(\log_3(x+4))=5$ gives $x\approx9.9024$ (from WolframAlpha)
So answer would be $x<9.9024$
Hint: take the logarithm w.r.t to $3$ on both sides of the inequality.