How do you solve #log_10y = (1/4)log_10(16) + (1/2)log_10(25)#?

1 Answer
Karthik G
Dec 21, 2015

#y = 10 #

Explanation:

#log_10 (y) = (1/4) log_10 (16) + (1/2) log_10(25) #

Rules which can be used here.

  1. #n*log(a) = log(a^n) #
  2. #log(a) + log(b) = log(ab)#
  3. If #log(a) = log(b)# then #a=b#

#log_10 (y) = log_10 (16)^(1/4) + log_10 (25)^(1/2)# By rule 1.
#log_10 (y) = log_10 (2) + log_10 (5) # since #a^(1/n) = root(n)a#
#log_10 (y) = log_10 (2*5)# By rule 2.
#log_10 (y) = log_10 (10) #

#y = 10# By Rule 3.

#y = 10 # is the final answer.

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