Simplify #2^(3+5log_2x)#?

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Answer 1 · 2018-02-08T14:51:39

#2^(3+5log_2x)=8x^5#

Let #a^(log_ax)=u#. Then taking logarithm to the base #a# on both sides we get

#log_ax xxlog_aa=log_au#

or #log_au=log_ax# and therefore #u=x# i.e.

#a^(log_ax)=x#

Using this in #2^(3+5log_2x)#

= #2^3xx2^(5log_2x)#

= #2^3xx(2^(log_2x))^5# - as #a^(mn)=(a^m)^n# or #(a^n)^m#

= #8x^5#