How do you solve #\log _{3}8\times \log _{8}9=2x#?

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Answer 1 · 2017-01-14T03:18:16

#x = 1#

This can be rewritten using #log_a n = logn/loga#.

#log8/log3 * log9/log8 = 2x#

#log9/log3 = 2x#

#log(3^2)/log(3^1) = 2x#

Use #loga^n = nloga#:

#(2log3)/log3 = 2x#

#2 = 2x#

#x = 1#

Hopefully this helps!