SOLUTION: i am having a hard time answering these problems. 1. x=8log(100,10) 2. logx (1/16)=-2 3. log(x)=square root log(x)

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: i am having a hard time answering these problems. 1. x=8log(100,10) 2. logx (1/16)=-2 3. log(x)=square root log(x)      Log On


   

Question 34026This question is from textbook college algebra
: i am having a hard time answering these problems.
1. x=8log(100,10)
2. logx (1/16)=-2
3. log(x)=square root log(x) This question is from textbook college algebra

Answer by xcentaur(357) About Me  (Show Source):
You can put this solution on YOUR website!
1. x=8log%28100%2C10%29
we know that
10%5E2=100OR100%5E%281%2F2%29=10
therefore,
log%28100%2C10%29=1%2F2
so we get,
x=8(1/2)=4
2. logx (1/16)=-2
take x to the other side,so we get
%281%2F16%29=x%5E%28-2%29
16%5E%28-1%29=x%5E%28-2%29
4%5E%282%29%5E%28-1%29=x%5E%28-2%29
4%5E%28-2%29=x%5E%28-2%29
therefore,x=4
3. log(x)=square root log(x)
square both sides,
[log(x)]^2=[log(x)]
using property of logs we get,
2logx=logx
2logx-logx=0
logx=0


now look at this,
log8 to the base 2=*3*
so 8=2 raised to *3*
instead of 3 if we put 0,instead of 8 we will get 1
so any number raised to zero becomes one,so we get
x=1


hope this helps,
-xC