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2015-03-24T00:06:10+05:30

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Log_a\ b=\frac{Log_{10}\ b}{log_{10}\ a}\\\\Log_{\frac{1}{3}}\ 7\ +\ 2\ Log_9\ 49\ -\ Log_{\sqrt3}\ \frac{1}{7}\\\\=\frac{Log_{10}\ 7}{Log_{10}\ \frac{1}{3}}+2\frac{Log\ 49}{Log\ 9}-\frac{Log\ \frac{1}{7}}{Log\sqrt3}\\\\=\frac{Log_{10}\ 7}{-Log_{10}\ 3}+2\frac{Log\ 7^2}{Log\ 3^2}-\frac{-Log\ 7}{\frac{1}{2}Log\ 3}\\\\=-\frac{Log_{10}\ 7}{Log_{10}\ 3}+2*\frac{2*Log_{10}\ 7}{2*Log\ 3}+\frac{2*Log\ 7}{Log\ 3}\\\\=3*\frac{Log_{10}\ 7}{Log_{10}\ 3}\\\\=3*Log_3\ 7
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this questions is not clear... really. Whether 4,81,25, 729 are bases or numbers ??

Log\ 4^{9\ log\ 81^{16\ Log\ 25^{27\ log\ 729^{25}}}}\\\\=9*log\ 81^{16\ Log\ 25^{27\ log\ 729^{25}}}*Log\ 4\\\\=9*(16\ Log\ 25^{27\ log\ 729^{25}})*(Log\ 81)*(2Log\ 2)\\\\=9*(16*(27\ log\ 729^{25}})*(Log\ 25)*(Log\ 81)*(2Log\ 2)\\\\=9*16*27*25*(Log\ 729)*(2*Log5)*(4Log3)*(2Log2)\\\\=9*16*27*25*(6Log3)*16*(Log5)(Log3)(Log2)\\\\=9*16*27*25*96*(Log5)(log3)^2*(log2)\\
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here i assume that for the above problem, 4, 81, 25, 729 are bases to the logarithms.

let\ x=Log_4\ (9\ Log_{81}\ (16 Log_{25}\ (27\ Log_{729}\ 25)))\\\\27*Log_{729}\ 25=27*\frac{Log25}{Log729}=27*\frac{2Log5}{6Log3}\\=\frac{9Log5}{Log3}\\\\16 Log_{25}\ (27\ Log_{729}\ 25)=16*\frac{Log9\frac{log5}{log3}}{Log25}=16*\frac{(Log9+Log(log5)-Log(log3))}{2log5}\\\\

It is getting complicated this way.  perhaps this is not the correct question..
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3)
x=2^{2-Log_2\ 5}\\\\Log_2\ x\\=2-Log_2\ 5\\=2*Log_2\ 2-Log_2\ 5\\=Log_2\ 2^2-Log_2\ 5\\=Log_2\ 2^2*5^{-1}\\=>x=4/5

1 5 1
sir why you took only 4, 81, 25, 729 as bases to the logarithms.
sir for first one no option