SOLUTION: 1.Given Loga2=x, loga3=y, loga5=z. Find loga60 2.Solve and find the product of the solutions 16^x=1/(4^-8/x) 3.find the Domain and range of y=log3(x-3)+10

Question 454754: 1.Given Loga2=x, loga3=y, loga5=z. Find loga60
2.Solve and find the product of the solutions 16^x=1/(4^-8/x)
3.find the Domain and range of y=log3(x-3)+10
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
1.Given Loga2=x, loga3=y, loga5=z. Find loga60
2.Solve and find the product of the solutions 16^x=1/(4^-8/x)
3.find the Domain and range of y=log3(x-3)+10
..
1.Given Loga2=x, loga3=y, loga5=z. Find loga60
loga(60)=loga(4)+loga(3)+loga(5)
loga(60)=loga(2^2)+loga(3)+loga(5)
loga(60)=2loga(2)+loga(3)+loga(5)
loga(60)=2x+y+z
..
2.Solve and find the product of the solutions 16^x=1/(4^-8/x)
16^x=1/(4^-8/x)
16^x=4^8/x)
4^2x=4^8/x
2x=8/x
2x^2=8
x^2=4
x=�2
Check:
16^x=16^2=256
4^8/x=4^4=256
..
3.find the Domain and range of y=log3(x-3)+10
y=log3(x-3)+10
Domain:
x-3>0
x>3
..
Range: all real numbers, (-∞,∞)
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