Question

If the value of log (0.25 to the base 4)+log(0.0625 to the base 2) is A,then find -1250A

Answer 1

The second solution is more exact.

$log_4\ 0.25 + log_2\ 0.0625 = A\\ \\Log_4\ 0.25 = x\\.\ \ \ 4^x = 0.25,\ \ \ (2^2)^x=\frac{25}{100},\ \ \ 2^{2x}=\frac{25}{100}\\. \ \ \ Log_{10}\ 2^{2x}=Log_{10}\ \frac{5^2}{10^2}\\.\ \ \ 2x\ Log\ 2=Log\ 5^2-Log\ 10^2= 2Log\ 5-2\\.\ \ \ \ x=\frac{Log_{10}5-1}{Log_{10}2}\\ \\log_2\ 0.0625 = y\\.\ \ \ 2^y=0.0625=0.25^2=0.5^4,\ \ \ Log_{10}2^y=Log_{10}0.5^4\\.\ \ y\ Log_{10}2=Log_{10}0.5^4,\ \ \ y = 4 Log_{10}(5/10)=4(Log_{10}5-Log_{10}10})\\.\ \ \ y=\frac{4(Log_{10}5-1)}{Log_{10}2}$


$A=x+y=\frac{Log_{10}5-1+4Log_{10}5-4}{Log_{10}2}=\frac{5Log_{10}5-5}{Log_{10}2}\\ \\-1250A=1250*(5-5Log_{10}5)/Log_{10}2=1250*1.505/0.3010\\ \\=6250.62$

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Uses the principle:  Log_x  A  =  Log_10   A  *  Log_x  10

$Log_{4}(25/100) = Log_{10} 0.25\ * Log_4 10 = \frac{Log_{10} 0.25}{Log_{10}4}=\frac{2Log_{10} 0.5}{2Log_{10}2}\\ \\=Log_2 0.5\\ \\A=Log_2 0.5+Log_2 0.0625=Log_2\ 0.5*0.25^2=Log_2\ 0.5^5=5*Log_20.5\\ \\A=5*Log_2\frac{1}{2}=5*(Log_2\ 1-Log_2\ 2)=5*(0-1)=-5\\ \\-1250*A = 6250$

Answer 2

Step-by-step explanation:

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