Question 1210375
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Let {{{log(5,(2)) = x}}} and {{{log(5,(3)) = y}}}


To figure out what {{{log(45,(100))}}} is in terms of x and y, we'll be using the Change of Base Rule


That rule is:
{{{log(b,(x)) = (log(c,(x)))/(log(c,(b)))}}}
where b is the original base and c is a new base to apply. 
We can select any positive real number for c as long as {{{c <> 1}}} 
If c = 1 was the case, then we'd have a division by zero error.


Since we're dealing with {{{log(45,(100))}}}, the original base is b = 45 and the input or argument to the log is x = 100.


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The variables x and y involve logs with base 5, so let's use c = 5.
{{{log(b,(x)) = (log(c,(x)))/(log(c,(b)))}}}


{{{log(45,(100)) = (log(5,(100)))/(log(5,(45)))}}}


{{{log(45,(100)) = (log(5,(2^2*5^2)))/(log(5,(3^2*5)))}}} Rewrite 100 as 2^2*5^2 and 45 as 3^2*5


{{{log(45,(100)) = (log(5,(2^2))+log(5,(5^2)))/(log(5,(3^2))+log(5,(5)))}}} Use the log rule log(A*B) = log(A)+log(B)


{{{log(45,(100)) = (2*log(5,(2))+2*log(5,(5)))/(2*log(5,(3))+log(5,(5)))}}} Applying log rule log(A^B) = B*log(A) to pull down the exponents.


{{{log(45,(100)) = (2*log(5,(2))+2*1)/(2*log(5,(3))+1)}}} When the log base and argument matches up, the result of the log is 1.


{{{log(45,(100)) = (2x+2)/(2y+1)}}} Apply the substitutions for x and y.
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