Find \(\large \log_2 3 \times \log_3 4 \times \log_4 5 \times \cdots\times \log_{127} 128 = \, ?\)
+23246 Using the change-of-base formula, writing each of these logs as logs with base 10:
log2(3) = log(3) / log(2)
log3(4) = log(4) / log(3)
log4(5) = log(5) / log(4)
...
log126(127) = log(127) / log(126)
log127(128) = log(128) / log(127)
When you multiply all of these together, notice that only log(3) from the first factor and log(128)
from the last factor remain, all the rest cancel out.
log(128) / log(2) = log( 27 ) / log( 2 ) = 7 · log( 2 ) / log( 2 ) = 7
