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MISCELLANEOUS EXERCISE 6 (II) · Q178

Q.Show that 7log⁡(1516)+6log⁡(83)+5log⁡(25)+log⁡(3225)=log⁡37\log\left(\dfrac{15}{16}\right)+6\log\left(\dfrac{8}{3}\right)+5\log\left(\dfrac{2}{5}\right)+\log\left(\dfrac{32}{25}\right)=\log3.

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7log⁡1516+6log⁡83+5log⁡25+log⁡3225=log⁡[(1516)7(83)6(25)53225]7\log\dfrac{15}{16}+6\log\dfrac{8}{3}+5\log\dfrac{2}{5}+\log\dfrac{32}{25}=\log\left[\left(\dfrac{15}{16}\right)^7\left(\dfrac{8}{3}\right)^6\left(\dfrac{2}{5}\right)^5\dfrac{32}{25}\right] (power rule then product rule).

Write every factor in primes: 15=3⋅5, 16=24, 8=23, 2=2, 5=5, 32=25, 25=5215=3\cdot5,\ 16=2^4,\ 8=2^3,\ 2=2,\ 5=5,\ 32=2^5,\ 25=5^2.

(1516)7=37⋅57228\left(\dfrac{15}{16}\right)^7=\dfrac{3^7\cdot5^7}{2^{28}}; (83)6=21836\left(\dfrac{8}{3}\right)^6=\dfrac{2^{18}}{3^6}; (25)5=2555\left(\dfrac{2}{5}\right)^5=\dfrac{2^5}{5^5}; 3225=2552\dfrac{32}{25}=\dfrac{2^5}{5^2}. …

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