SOLUTION: I'm trying to convert this equation: log base 100 (1/100x). I got to Log base 100 (1) - log base 100 (100x) but I don't know the missing steps to get to: -1-log (sqrt of x)

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I'm trying to convert this equation: log base 100 (1/100x). I got to Log base 100 (1) - log base 100 (100x) but I don't know the missing steps to get to: -1-log (sqrt of x)      Log On


   



Question 270396: I'm trying to convert this equation: log base 100 (1/100x). I got to Log base 100 (1) - log base 100 (100x) but I don't know the missing steps to get to: -1-log (sqrt of x)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%28100%2C+%281%2F%28100x%29%29%29+=+log%28100%2C+%281%29%29+-+log%28100%2C+%28100x%29%29
You are correct so far. To continue we can first simplify the first logarithm. Since 100%5E0+=+1 log%28100%2C+%281%29%29+=+0. So now we have
0+-+log%28100%2C+%28100x%29%29+=+-log%28100%2C+%28100x%29%29
Now we can use the property of logarithms, log%28a%2C+%28p%2Aq%29%29+=+log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29 to separate the remaining log into two:
-log%28100%2C+%28100x%29%29+=+-%28log%28100%2C+%28100%29%29+%2B+log%28100%2C+%28x%29%29%29
Note the use of parentheses. Whenever you substitute one expression for another, especially when there are a different number of terms, it is a very good idea to use parentheses like this. In this case, it helps us understand that the "-" in front applies to both terms.
The first log is 1 so

At this point we have the -1 part of the answer. The other term in the answer is a base 10 logarithm, not a base 100 logarithm. So we need to change the base of our remaining logarithm using the formula: log%28a%2C+%28p%29%29+=+log%28b%2C+%28p%29%29%2Flog%28b%2C+%28a%29%29. Using this to change our base 100 logarithm into a base 10 logarithm we get:
-1+-+log%28100%2C+%28x%29%29+=+-1+-+log%28%28x%29%29%2Flog%28%28100%29%29
And since 10%5E2+=+100 log%28%28100%29%29+=+2:
-1+-+log%28%28x%29%29%2Flog%28%28100%29%29+=+-1+-+log%28%28x%29%29%2F2
We're getting close. We have the -1 and we have the base 10 logarithm. We only have to figure out how log%28%28x%29%29%2F2+=+log%28%28sqrt%28x%29%29%29. Since dividing by 2 is the same as multiplying by 1/2 we can rewrite the term as a multiplication:
-1+-+log%28%28x%29%29%2F2+=+-1+-+%281%2F2%29log%28%28x%29%29
Now we can use the property of logarithms, q%2Alog%28a%2C+%28p%29%29+=+log%28a%2C+%28p%5Eq%29%29 to move the 1/2 into the argument as an exponent:
-1+-+%281%2F2%29log%28%28x%29%29+=+-1+-+log%28%28x%5E%281%2F2%29%29%29
And since an exponent of 1/2 means square root:
-1+-+log%28%28x%5E%281%2F2%29%29%29+=+-1+-+log%28%28sqrt%28x%29%29%29
And we're done!