SOLUTION: I am having some issues with logarithm equations and need some help. first problem: 2 log(z)-log(z2+4z+1)=0 The 2 after the z here is z squared but I can't get a little 2 on m

Question 660356: I am having some issues with logarithm equations and need some help.
first problem: 2 log(z)-log(z2+4z+1)=0 The 2 after the z here is z squared but I can't get a little 2 on my pc.
next problem: 2e3y+8 - 11e5-10y = 0 here the 3y+8 is to the power and small again and so is the 5-10y but i can't get it on my pc.
next problem: 1-8In {_2x-1___} = 14 That is 2x - 1 over 7
7
I am an a student usually, but we changed math teachers 3 x so far this year and this is causing issues with learniing. This is hwork. We had 30 problems. I did what I could and am stumped on about 5 or 6 of them. Here are 3...If i can get these, I can probably figure out the others. I have been up all nite with this and just went to bed and will be getting up at 5 to finish, so if anyone up, I would really appreciate the help. Thank you much.
Bob
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
first problem: 2 log(z)-log(z2+4z+1)=0 The 2 after the z here is z squared but I can't get a little 2 on my pc.
2 log(z)-log(z2+4z+1)=0
log(z^2)=log(z^2+4z+1)
z^2=z^2+4z+1
4z+1=0
z=-1/4
no solution: z>0
..
next problem: 2e3y+8 - 11e5-10y = 0 here the 3y+8 is to the power and small again and so is the 5-10y but i can't get it on my pc.
2e^(3y+8) - 11e^(5-10y) = 0
2e^(3y+8) = 11e^(5-10y)
take log of both sides
ln2+(3y+8)lne=ln11+(5-10y)lne
lne=1
ln2+3y+8=ln11+5-10y
3y+8-5+10y=ln11-ln2
13y+3=(ln11-ln2)
y=((ln11-ln2-3)/13≈-.0996
..
next problem: 1-8In {_2x-1___} = 14 That is 2x - 1 over 7
1-8ln[(2x-1)/7]=14 (I hope I got this right)
-8ln[(2x-1)/7]=13
ln[(2x-1)/7]=-13/8
convert to exponential form:
e^(-13/8)=(2x-1)/7
using calculator:
.1969=(2x-1)/7
2x-1=.1969*7=1.3783
2x=2.3783
x=2.3783/2=1.1892
note: answers are rounded to four decimal places
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