SOLUTION: Hi, I'm stuck on the following problem. Would you help me solve it, please? Directions: Solve the equation and check all proposed answers. {{{ (x-13)^(2/5)=9 }}} I got 256

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hi, I'm stuck on the following problem. Would you help me solve it, please? Directions: Solve the equation and check all proposed answers. {{{ (x-13)^(2/5)=9 }}} I got 256       Log On


   



Question 906529: Hi, I'm stuck on the following problem. Would you help me solve it, please?
Directions: Solve the equation and check all proposed answers.
+%28x-13%29%5E%282%2F5%29=9+
I got 256 as an answer by raising both sides to the (5/2) power, but my online homework system is saying that's only 50% correct. Did I forget another answer or important step?

Found 3 solutions by Edwin McCravy, mananth, josgarithmetic:
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, I'm stuck on the following problem. Would you help me solve it, please?
Directions: Solve the equation and check all proposed answers.
matrix%282%2C1%2C%22%22%2C+%28x-13%29%5E%282%2F5%29=9%29+
I got 256 as an answer by raising both sides to the (5/2) power, but my online homework system is saying that's only 50% correct. Did I forget another answer or important step?
Don't raise both sides to the 'reciprocal power' ever again. Raise both sides to
the 'denominator power' instead. I know you were taught to raise both sides to
the 'reciprocal power' when you first started taking algebra.  But don't
do that anymore because it sometimes causes you to lose solutions in more
advanced problems like this one which is more advanced than the ones you had
when you learned to use the reciprocal power. Raise both sides to the
denominator power, (the fifth power) not the reciprocal power.

matrix%282%2C1%2C%22%22%2C+%28%28x-13%29%5E%282%2F5%29%29%5E5=9%5E5%29+

%28x-13%29%5E2=9%5E5

Now use the principle of square roots:

x-13=%22%22%2B-sqrt%289%5E5%29

Write 9 as 3²

x-13=%22%22%2B-sqrt%28%283%5E2%29%5E5%29

Multiply the exponents

x-13=%22%22+%2B-+sqrt%283%5E10%29

x-13=%22%22+%2B-+3%5E5

x-13=%22%22+%2B-+243

x=13+%2B-+243

So there are 2 solutions.

Using the +,

x=13%2B243

x=256

Using the -

x=13-243

x=-230

That's the other answer.  Don't raise both sides to the reciprocal power
ever again.  Just raise to the denominator power. 

Edwin

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
+%28x-13%29%5E%282%2F5%29=9+

+%28%28x-13%29%5E%282%2F5%29%29%5E5=%283%5E2%29%5E5+

%28x-13%29%5E2=3%5E10
take square root
x-13 = +/- 3^5
x-13= +/- 243
x= 13+243 or 13-243
x=256 OR -230

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
%28x-13%29%5E%282%2F5%29=9
x-13=0%2B-9%5E%285%2F2%29, because this includes a half-power, same as square root.
You are doing two things: Finding/taking square root of 9, and then raising
that result to the 5th power; the raise to the 1%2F2 power brings up PLUS-or-MINUS,
but the raise to 5th power is not sensitive the the sign.