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Algebra Level 3

A line, x=c intersect both the graphs of y=logx/log5 and y=log(x+4)/log5. If the distance between the points of intersection is 1/2, and if c can be expressed as a+√b where a and b are relatively coprime integers and b is square free, compute 100a+b.


The answer is 105.

Jun Arro Estrella by Jun Arro Estrella , 23, Philippines

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2 solutions

Jun Arro Estrella
Dec 24, 2015

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Parveen Soni
Oct 27, 2014

say graph 1 is y1 and graph 2 is y2 where
y1=log c/log 5 when x=c
and y2=[log (c+4)]/log 5 when x=c
and y2-y1= [(log c+4)-log c]/log 5=(log c+4/c ) /log 5
Two point of intersection will be (c,y1) and (c,y2)
now distance between two point(on same straight line) is
1/2={(y2-y1)^2+(c-c)^2}^1/2 (using distance formulae)
which gives
y2-y1=1/2 i.e (log c+4/c)/log 5=1/2
now using log properties we have
{(c+4)/c}^2=5 i.e c^2-2c-4=0
now roots are c1=1+(5^1/2)
and c2=1-(5^1/2).
Since a and b are relatively prime(always +ve)
Therefore reject root c2
comparing c1 with standard given we have
a=1 and b=5 which means 100a+b=105



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