If #5^(10x)=4900# and #2^(sqrty)=25#, what is the value of #((5^((x-1)))^5)/4^(-sqrty)#?
1 Answer
Mar 4, 2017
Explanation:
Given that:
#5^(10x) = 4900#
#2^(sqrt(y)) = 25#
Then:
#(5^(x-1))^5 = 5^(5(x-1))#
#color(white)((5^(x-1))^5) = 5^(5x-5)#
#color(white)((5^(x-1))^5) = 5^(5x)*5^(-5)#
#color(white)((5^(x-1))^5) = 5^(1/2(10x))*5^(-5)#
#color(white)((5^(x-1))^5) = (5^(10x))^(1/2)*5^(-5)#
#color(white)((5^(x-1))^5) = 4900^(1/2)*5^(-5)#
#color(white)((5^(x-1))^5) = 70*5^(-5)#
#color(white)((5^(x-1))^5) = 14*5^(-4)#
#4^(-sqrt(y)) = (2^2)^(-sqrt(y))#
#color(white)(4^(-sqrt(y))) = 2^(-2sqrt(y))#
#color(white)(4^(-sqrt(y))) = (2^sqrt(y))^(-2)#
#color(white)(4^(-sqrt(y))) = 25^(-2)#
#color(white)(4^(-sqrt(y))) = 5^(-4)#
So:
#(5^(x-1))^5/4^(-sqrt(y)) = (14*color(red)(cancel(color(black)(5^(-4)))))/color(red)(cancel(color(black)(5^(-4)))) = 14#