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Algebra Level 3

Evaluate

log ⁡ 2 x + log ⁡ 4 x + log ⁡ 16 x + log ⁡ 256 x + log ⁡ 65536 x + ⋯ . \log_ 2 x + \log_4 x + \log_{16} x + \log_{256} x + \log_{65536} x + \cdots \; .

Note that the successive subscripts are squares of each other - 2 2 = 4 , 4 2 = 16 , 1 6 2 = 256 , 25 6 2 = 65536 2^2 = 4, 4^2 = 16, 16^2 = 256, 256^2 = 65536 .

log ⁡ 2 ( 2 x ) \log_2 (2x) [ log ⁡ 2 x ] 2 [\log_2 x ]^2 [ log ⁡ 2 ( 2 x ) ] 2 [\log_2 (2x) ]^2 log ⁡ 2 ( x 2 ) \log_2 ( x^2 )

Calvin Lin by Calvin Lin

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2 solutions

Just changed the base of every log so it has en base 2. You will get log ⁡ 2 ( x ) ( 1 + 1 2 + 1 4 + … ) = 2 log ⁡ 2 ( x ) = log ⁡ 2 ( x 2 ) \log_2 (x) (1 + \frac {1}{2} + \frac {1}{4} + \ldots) = 2 \log_2 (x) = \log_2 (x^2)

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Note: To type in Latex, you just need to add the latex brackets of \ ( \ ) \backslash ( \quad \backslash) . I've edited the solution for you. You can click on the edit button to see what I did.

Calvin Lin Staff - 4 years, 7 months ago

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Tnx... never drink and latex ;)

Peter van der Linden - 4 years, 7 months ago

log ⁡ 2 x + log ⁡ 4 x + log ⁡ 16 x + log ⁡ 256 x ⋯ = log ⁡ 2 x + 1 2 log ⁡ 2 x + 1 4 log ⁡ 2 x + 1 8 log ⁡ 2 x + 1 16 log ⁡ 2 x ⋯ \log_{2}x+\log_{4}x+\log_{16}x+\log_{256}x \cdots =\log_{2}x+\dfrac{1}{2}\log_{2}x+\dfrac{1}{4}\log_{2}x+\dfrac{1}{8}\log_{2}x+\dfrac{1}{16}\log_{2}x \cdots = log ⁡ 2 x × ( 1 + 1 2 + 1 4 + 1 8 ⋯ ) = log ⁡ 2 x × ( 1 1 − 1 2 ) = 2 log ⁡ 2 x = log ⁡ 2 ( x 2 ) =\log_{2}x\times (1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8} \cdots)=\log_{2}x\times(\dfrac{1}{1-\dfrac{1}{2}}) =2\log_{2}x=\color{#D61F06}{\boxed{\log_{2}(x^2)}}

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