Question #4e300

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Answer 1 · 2017-12-01T10:51:22

# x=1/2, or, x=2.#

Using the Change of Base Rule with base #e,# we have

#log_e 2/log_ex=4(log_e x/log_e 16).#

#:. 4(log_ex)^2=(log_e2)(log_e 16),#

#=(log_e2)(log_e 2^4),#

#=(log_e2)(4log_e2).#

#rArr 4(log_ex)^2=4(log_e2)^2.#

# rArr 4(log_ex)^2-4(log_e2)^2=0.#

# rArr 4{(log_ex)^2-(log_e2)^2}=0.#

#rArr{(log_ex)+(log_e2)}{(log_ex)-(log_e2)}=0.#

#rArrlog_ex+log_e2=0, or, log_2x-log_e2=0.#

#rArrlog_ex=-log_e2, or, log_2x=log_e2.#

#rArrlog_ex=log_e2^-1, or, log_2x=log_e2.#

Since, #log# function is #1-1,# we get,

#x=2^-1=1/2, or, x=2.#

Enjoy Maths.!