Aspire Faculty ID #16175 · Topic: NIMCET 2011 · Just now
NIMCET 2011

Solve inequality $\log_3\big((x+2)(x+4)\big)+\log_{1/3}(x+2)<\dfrac12\log_{\sqrt{3}}7$

Solution

Domain: $x>-2$ and $x>-4$ ⇒ $x>-2$. Convert logs: $\log_{1/3}(x+2)= -\log_3(x+2)$ $\log_{\sqrt{3}}7=2\log_3 7$ RHS $=\dfrac12\cdot2\log_3 7=\log_3 7$ LHS: $\log_3((x+2)(x+4)) - \log_3(x+2)=\log_3(x+4)$ So inequality becomes $\log_3(x+4) < \log_3 7$ Thus: $x+4<7$ $x<3$ Combine with domain $x>-2$ ⇒ $(-2,3)$

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