How do you solve #16^(5x)= 64^(x+7)#?

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Answer 1 · 2015-07-19T13:47:44

#color(green)(x=3#

expressing the bases in terms of #4#

#16=4^2#

#64=4^3#

Rewriting the expression:
#16^(5x)=64^(x+7)#

#4^(2*(5x))=4^(3*(x+7))#

#4^(10x)=4^(3x+21)#

Now we can equate the powers:

#10x=3x+21#

#7x=21#

#color(green)(x=3#