How do you solve #4^(2x+5) = 8^(3x)#?

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Answer 1 · 2016-05-17T10:06:56

#x=2#

#4^(2x+5)=8^(3x)#

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

#(a^m)^n=a^(m*n)#

#2^(4x+10)=2^(9x)#

#"Both side of equation have the same base value"#

#4x+10=9x#

#10=9x-4x#

#10=5x#

#x=10/5#

#x=2#