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)

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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

Found 2 solutions by xcentaur, MathTherapy:
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


Answer by MathTherapy(10699) About Me  (Show Source):
You can put this solution on YOUR website!
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)

The following has 2 solutions, not 1 (ONE)!
3. log(x)=square root log(x)
            log+%28%28x%29%29+=+sqrt%28log+%28%28x%29%29%29
        %28log+%28%28x%29%29%29%5E2+=+%28sqrt%28log+%28%28x%29%29%29%29%5E2 ----- Squaring both sides
        %28log+%28%28x%29%29%29%5E2+=+log+%28%28x%29%29 
%28log+%28%28x%29%29%29%5E2+-+log+%28%28x%29%29+=+0 
log (x)[log (x) - 1] = 0 ----- Factoring out GCF, log (x)
log (x) = 0        OR        log (x) - 1 = 0
log (x) = 0        OR        log (x) = 1
Converting to EXPONENTIAL form, we get:
highlight%28x%29+=+10%5E0+=+highlight%281%29    OR   highlight%28x%29+=+10%5E1+=+highlight%2810%29