How do you solve #1/2(log_7x+log_7 8)=log_7 16#?

1 Answer
Noah G
Nov 26, 2016

#log_7 x + log_7 8 = log_7 16/(1/2)#

#log_7 x + log_7 8 = 2log_7 16#

We now apply the rules #log_a n + log_a m = log_a (n xx m)# and #alogn = logn^a#.

#log_7(x xx 8) = log_7 16^2#

#log_7 8x = log_7 256#

If #loga = logb#, then #a = b#.

#8x = 256#

#x = 32#

Hopefully this helps!

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