How do you solve #log_2 x+log_4 x=log_2 5#?

1 Answer
De Rono
Nov 2, 2015

#x = 5^(2/3)#

Explanation:

Note that

# log _2 x+log_4x = log_2 5#
#implies (lnx)/(ln2) + (lnx)/(ln2^2) = (ln5)/(ln2)#
#implies (lnx)/(ln2) + (lnx)/(2ln2) = (ln5)/(ln2)#
#implies(lnx)/(ln2)(1+1/2) = (ln5)/(ln2)#
#implies lnx = 2/3 ln5#
#implies lnx = ln5^(2/3)#
#implies x = 5^(2/3)#

Related questions
See all questions in Logarithmic Models
Impact of this question
1722 views around the world
You can reuse this answer
Creative Commons License