How do you solve #8^y = 4^(2x+3)# and #log2y=log2x+4#?

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Answer 1 · 2016-04-30T04:41:52

The first equation becomes

#8^y=4^(2x+3)=>(2^3)^y=2^(2*(2x+3))=>3*y=4x+6#

The second equation becomes

#log2y=log2x+4=>log((2y)/(2x))=4=>log(y/x)=4=> y/x=10^4=>y=10^4*x#

Now the system of equations become

#3y=4x+6#
#y=10^4*x#

which has solutions

#x = 3/14998#, #y = 15000/7499#