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Find \(log_23\cdot log_34...log_(127)128\)

 Nov 7, 2022
 #1
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By the way, the (127) was supposed to be the bottom number, but the latex wouldn't allow more than one digit.

 Nov 7, 2022
 #4
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You needed to put the 127 in parenthesis, not brackets, they are the curly ones  {127 }

Melody  Nov 7, 2022
 #3
avatar+118135 
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\(log_23\cdot log_34...log_{127}128\\ =\frac{log3}{log2}\cdot \frac{log4}{log3}\dots \cdot \frac{log128}{log127}\\ cancel\\ =\frac{log128}{log2}\\ =\frac{log2^7}{log2}\\ =\frac{7log2}{log2}\\ =7\)

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 Nov 7, 2022
 #5
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thank you!

 Nov 8, 2022

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