How do you solve #x^(4/3)+9=25#?

1 Answer
Jan 9, 2016

Answer:

#x=8#

Explanation:

Firstly, we rearrange the equation so that there will only be equation with one unknown on one side while integer on the other side.

So,
#x^(4/3)+9=25#

#x^(4/3)=25-9#

#x^(4/3)=16#

To eliminate the power on unknown #x#, we need to whether add to the power of or to the power root of on both side of the equation. For example;

#x^(4/3)=16# also equals to #root(3)(x)^(4)=16#

Let say we want to eliminate the power root of #3# on #x#, we must add to the power of #3# onto both side of equation;

#x^(4/3)=16#
Add to the power of #3# on both side of equation;

#x^((4/3)(3))=16^3#

#x^(4)=4096#

Let say then we want to eliminate the power of #4# on #x#, we must add to the power root of #4# onto both side of equation;

#x^(4)=4096#
Add to the power root of #4# on both side of equation;

#x^((4)(1/4))=4096^(1/4)#

#x=8#