How do you solve #4^ { (2z) / 3} = 8^ { ( z + 2) }#?

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Answer 1 · 2016-11-10T16:35:42

#:.z=-(18)/5#

reduce the base numbers to the same value.

#4=2^2# and #2^3=8#

so

#4^((2z)/3)=8^(z+2)#

becomes

#(2^2)^((2z)/3)=(2^3)^(z+2)#

#2^((4z)/3)=2^(3(z+2))#

#=>(4z)/3=3z+6#

#(4z)/3-3z=6#

#-(5z)/3=6#

#:.z=-(18)/5#