The joint solution of inequality $\displaystyle \log_\pi (x+1)+\log_\pi x<\log_\pi (2x+6)$ is? Answer: $\displaystyle \{x \in IR;0<x<3\}$
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Originally Posted by boy47 The joint solution of inequality $\displaystyle \log_\pi (x+1)+\log_\pi x<\log_\pi (2x+6)$ is? Answer: $\displaystyle \{x \in IR;0<x<3\}$ Using the usual log rule and a bit of re-arranging you should get $\displaystyle x^2 - x - 6 < 0$ which has the solution $\displaystyle -2 < x < 3$. But $\displaystyle \log_{\pi} x$ is only defined for x > 0. Therefore ....
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