How do you solve #log_4x=log_8(4x)#?

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Answer 1 · 2016-08-23T09:24:24

#x = 2^4#

Using

#log_a b= log_e a/(log_e b)#

#log_4x = (log_e x)/(log_e 4) = log_8(4x) = (log_e (4x))/(log_e 8)#

but #log_e 4 = 2 log_e 2# and #log_e 8 = 3 log_e 2# so

#log_e x/(2log_e 2) = (log_e x + 2log_e 2)/(3 log_e 2)#

#3log_e x = 2log_e x+4 log_e 2#

and finally

#log_e x = log_e 2^4#

so

#x = 2^4#