How do you solve #(2^6 times 2^3)/(2^n) = 2^5#?

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Answer 1 · 2017-05-25T11:45:04

#n=4#

First multiply both sides by #color(green)2^color(green)n# :

#(2^6 times 2^3)/(2^n)# #times# #color(green)2^color(green)n# #=# #2^5# #times# #color(green)2^color(green)n#

This will remove #2^n# from the LHS :

#(2^6 times 2^3)/cancel(2^n)# #times# #cancel(2^n)# #=# #2^5# #times# #color(green)2^color(green)n#

Second add the exponents of similar bases :

#2^(6+3)##=# #2^(5+n)#

#=># #2^9##=# #2^(5+n)#

Third use the #"Base Rule"# :

i.e. #9= 5 + n#

#:.# #n=4#