SOLUTION: Hi i am having trouble with some of my trigonometry problems i dont get it at all it is way over my head 1] log(b) square root of x/y^2z 2] 5^x+1=41 3] log(1/2)8=-3 4] log(3)

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Question 32124: Hi i am having trouble with some of my trigonometry problems i dont get it at all it is way over my head
1] log(b) square root of x/y^2z
2] 5^x+1=41
3] log(1/2)8=-3
4] log(3) 81=x
i hope you answer this i really need help i dont understand this
thank you by the way for the last set of questions you helped me a lot
thank you
brittny
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
1)=(1/2)[logx-log(y^2z)]
= (1/2)[logx-[2logy +logz]]
=(1/2)[logx-2logy-logz]
2)5^x=40
x(log5)=log40
x=[log40/log5]
x=2.292...
3)log(1/2)8=-3
Yes, this is true as the following shows:
(1/2)^-3=8
4) log(3)81=x
3^x=81
3^x=3^4
x=4
OR
4)log(3)81=x
x=[log81/log3]
x=[1.908485019...]/[0.477121255...]=4
Cheers,
Stan H.