How do you solve #(4x + 8) ^ { 1/ 4} + 7= 9#?

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Answer 1 · 2016-10-19T20:03:40

#x = 2#

#(4x+8)^(1/4) +7 = 9" "larr# subtract 7 from both sides

#(4x+8)^(1/4) = 2#

This means the same as #" "root4 (4x+8) = 2#

Raise both sides to the power of 4

#root4 (4x+8)^4 = 2^4#

#4x +8 = 16#

#4x = 8#

#x = 2#