Question 537725
{{{(3/log(2,a))-(2/log(4,a))=1/log(1/2,a)}}}<P>
Step 1 is convert all the logs to the same base.  Use log base 10 because if you were doing this for a final calculation, you'd use log base 10.  Most calculators have a log (base 10) button.<P>
{{{log(2,a)=log(a)/log(2)}}}<P>
{{{log(4,a)=log(a)/log(4)}}}<P>
{{{log(1/2,a)=log(a)/log(1/2)}}}<P>
{{{(3/(log(a)/log(2)))-(2/(log(a)/log(4)))=(1/(log(a)/log(1/2)))}}}<P>
a/b/c = a*c/b<P>
When there's a fraction in the denominator, flip it and multiply it by the numerator.<P>
{{{3/(log(a)/log(2))=3*(log(2)/log(a))}}}<P>
{{{2/(log(a)/log(4))=2*(log(4)/log(a))}}}<P>
{{{1/(log(a)/log(1/2))=1*(log(1/2)/log(a))}}}<P>
That manipulation gives<P>
{{{3*(log(2)/log(a))-2*(log(4)/log(a))=1*(log(1/2)/log(a))}}}<P>
Remember that if {{{(a/b) + (c/b) = (d/b)}}} then {{{a+c=d}}}<P>
Drop the denominator log(a)<P>
{{{3*log(2)-2*log(4)=log(1/2)}}}<P>
Power rule is {{{x*log(y) = log(y^x)}}}<P>
{{{log(2^3) - log(4^2) = log (1/2)}}}<P>
Quotient rule is {{{log(x)-log(y) = log(x/y)}}}<P>
{{{log((2^3)/(4^2)) = log(1/2)}}}<P>
{{{2^3=8}}} and {{{4^2=16}}}<P>
{{{log(8/16)=log(1/2)}}}<P>
Divide 8 and 16 by 8.<P>
{{{log(1/2) = log(1/2)}}}<P>
As the Fonz would say, "Checkamundo!"
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