Question 32124
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.