Good for 15 seconds question! (Part 5)

Algebra Level 3

Find the value of: ( l o g 3 4 ) ( l o g 4 5 ) ( l o g 5 6 ) ( l o g 6 7 ) … ( l o g 80 81 ) (log_3 4)(log_4 5)(log_5 6)(log_6 7)\ldots(log_{80} 81)

See part 4 here


The answer is 4.

Mj Santos by MJ Santos , 22, Philippines

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3 solutions

Mj Santos
Feb 9, 2015

Note that: ( l o g 3 4 ) ( l o g 4 5 ) ( l o g 5 6 ) ( l o g 6 7 ) … ( l o g 80 81 ) = ( l o g 4 l o g 3 ) ( l o g 5 l o g 4 ) … ( l o g 81 l o g 80 ) = l o g 81 l o g 3 = l o g 3 81 = 4 (log_3 4)(log_4 5)(log_5 6)(log_6 7)\ldots(log_{80} 81)=(\frac{log4}{log3})(\frac{log5}{log4})\ldots(\frac{log81}{log80})=\frac{log81}{log3}=log_3 81=\boxed4

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log ⁡ a b = log ⁡ b log ⁡ a \log_{a} b = \dfrac{ \log b}{\log a}

sandeep Rathod - 6 years, 4 months ago

15 secs is toooo much... it needs max of 5 secs

Md Zuhair - 3 years, 5 months ago
Fox To-ong
Feb 12, 2015

cancelling all leading to log of 81 to the base of 3 = 4

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Hajar Aggad
Feb 22, 2015

Loga(b) = ln(b)/ln(a) so we gonna simplify until we get : ln(81)/ln(3)
and that's equal to : log3(81)= 4

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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