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)

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) (Show Source): You can put this solution on YOUR website!
1.
we know that
OR
therefore,

so we get,
x=8(1/2)=4
2. logx (1/16)=-2
take x to the other side,so we get

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

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