How do you solve #{3x-2)^(3/5) = 8 #?

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Answer 1 · 2016-04-26T18:11:02

#x=34/3#

Our first step should be to undo the #3/5# power. In order to do this, raise each side to the #5/3# power.

#((3x-2)^(3/5))^(5/3)=8^(5/3)#

On the left hand side, use the following rule:

#(a^b)^c=a^(bc)#

Applying this rule to the left side, we multiply #3/5# and #5/3#:

#(3x-2)^(3/5xx5/3)=8^(5/3)#

#(3x-2)^1=8^(5/3)#

#3x-2=8^(5/3)#

Now, we should simplify #8^(5/3)#. Do this by writing #8# as #2^3#.

#3x-2=(2^3)^(5/3)#

Use the exponent rule again.

#3x-2=2^((3xx5/3))#

#3x-2=2^5#

#3x-2=32#

#3x=34#

#x=34/3#