BITSAT 2019 · Mathematics · Basics of Mathematics
C. x>323x > \dfrac{3}{23}x>233
We have, x=log53+log75+log97x = \log_{5}3 + \log_{7}5 + \log_{9}7x=log53+log75+log97Using AM≥GMAM \ge GMAM≥GM for three positive real numbers a+b+c3≥(abc)13\dfrac{a + b + c}{3} \ge \left(abc\right)\dfrac{1}{3}3a+b+c≥(abc)31. Equality holds when a=b=c.a = b = c.a=b=c.∴log53+log75+log973>(log53⋅log75⋅log97)1/3⇒x>3(log3log5×log5log7×log7log9)1/3∴ \dfrac{\log_{5}3 + \log_{7}5 + \log_{9}7}{3} > \left(\log_{5}3 \cdot \log_{7}5 \cdot \log_{9}7\right)^{1} / 3⇒ x > 3\left(\dfrac{\log 3}{\log 5} \times \dfrac{\log 5}{\log 7} \times \dfrac{\log 7}{\log 9}\right)^{1} / 3∴3log53+log75+log97>(log53⋅log75⋅log97)1/3⇒x>3(log5log3×log7log5×log9log7)1/3⇒x>3(12)1/3⇒x>323⇒ x > 3\left(\dfrac{1}{2}\right)^{1} / 3 ⇒ x > \dfrac{3}{23}⇒x>3(21)1/3⇒x>233
We have, x=log53+log75+log97x = \log_{5}3 + \log_{7}5 + \log_{9}7x=log53+log75+log97
Using AM≥GMAM \ge GMAM≥GM for three positive real numbers a+b+c3≥(abc)13\dfrac{a + b + c}{3} \ge \left(abc\right)\dfrac{1}{3}3a+b+c≥(abc)31. Equality holds when a=b=c.a = b = c.a=b=c.
∴log53+log75+log973>(log53⋅log75⋅log97)1/3⇒x>3(log3log5×log5log7×log7log9)1/3∴ \dfrac{\log_{5}3 + \log_{7}5 + \log_{9}7}{3} > \left(\log_{5}3 \cdot \log_{7}5 \cdot \log_{9}7\right)^{1} / 3⇒ x > 3\left(\dfrac{\log 3}{\log 5} \times \dfrac{\log 5}{\log 7} \times \dfrac{\log 7}{\log 9}\right)^{1} / 3∴3log53+log75+log97>(log53⋅log75⋅log97)1/3⇒x>3(log5log3×log7log5×log9log7)1/3
⇒x>3(12)1/3⇒x>323⇒ x > 3\left(\dfrac{1}{2}\right)^{1} / 3 ⇒ x > \dfrac{3}{23}⇒x>3(21)1/3⇒x>233
