Question 270396
{{{log(100, (1/(100x))) = log(100, (1)) - log(100, (100x))}}}
You are correct so far. To continue we can first simplify the first logarithm. Since {{{100^0 = 1}}} {{{log(100, (1)) = 0}}}. So now we have
{{{0 - log(100, (100x)) = -log(100, (100x))}}}
Now we can use the property of logarithms, {{{log(a, (p*q)) = log(a, (p)) + log(a, (q))}}} to separate the remaining log into two:
{{{-log(100, (100x)) = -(log(100, (100)) + log(100, (x)))}}}
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 <i>to both terms</i>.
The first log is 1 so
{{{-(log(100, (100)) + log(100, (x))) = -(1 + log(100, (x))) = -1 - log(100, (x))}}}
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(a, (p)) = log(b, (p))/log(b, (a))}}}. Using this to change our base 100 logarithm into a base 10 logarithm we get:
{{{-1 - log(100, (x)) = -1 - log((x))/log((100))}}}
And since {{{10^2 = 100}}} {{{log((100)) = 2}}}:
{{{-1 - log((x))/log((100)) = -1 - log((x))/2}}}
We're getting close. We have the -1 and we have the base 10 logarithm. We only have to figure out how {{{log((x))/2 = log((sqrt(x)))}}}. Since dividing by 2 is the same as multiplying by 1/2 we can rewrite the term as a multiplication:
{{{-1 - log((x))/2 = -1 - (1/2)log((x))}}}
Now we can use the property of logarithms, {{{q*log(a, (p)) = log(a, (p^q))}}} to move the 1/2 into the argument as an exponent:
{{{-1 - (1/2)log((x)) = -1 - log((x^(1/2)))}}}
And since an exponent of 1/2 means square root:
{{{-1 - log((x^(1/2))) = -1 - log((sqrt(x)))}}}
And we're done!