How do you solve #5root3(4x-1)+2=17#?

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Answer 1 · 2016-06-01T06:43:32

#x=7#

#5xxroot(3)(4x-1)+2=17#

Here, we try to keep the cube root term on one side.

Subtract #2# from both sides:

#5xxroot(3)(4x-1)+2color(red)(-2)=17color(red)(-2)#

#5xxroot(3)(4x-1)=15#

Next, divide both sides by #5#:

#[5xxroot(3)(4x-1)]/color(red)(5)=15/color(red)(5)#

#root(3)(4x-1)=3#

Now, we remove the cube root.
In order to do so, find the cube of both sides:

#(root(3)(4x-1))^color(red)(3)=3^color(red)(3)#

#4x-1=3xx3xx3#

#4x-1=27#

#4x-1color(red)(+1)=27color(red)(+1)#

#4x=28#

#(4x)/color(red)(4)=28/color(red)(4)#

#x=7#