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Grade 12Algebra

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Profile image of Aman Gandhi
11 Years agoGrade 12
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1 Answer

Profile image of Nishant Vora
11 Years ago
Hello Student, Please find the solution
A= log m + log n
=> A= log (mn)
=>A= \log_{10} (\frac{(ab)^2 - [(ab)^2 - 4(a+b)]}{4})

A= \log_{10} [\frac{ 4(a+b)}{4}]

A= \log_{10} [\frac{ 4(43+57)}{4}] = \log_{10} 100 = 2

Now, B= 2^{\log_{6} 18} 3^{\log_{6} 3}

B= 2^{\log_{6} 6*3} 3^{\log_{6} 3}

B= 2^{\log_{6} 6+ \log_{6}3} 3^{\log_{6} 3}
B= 2^{1+ \log_{6}3} 3^{\log_{6} 3}
B= 2*2^{\log_{6}3} 3^{\log_{6} 3}
B= 2*6^{\log_{6}3}
B= 2*3 =6

So A*B = 6*2 = 12


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