(x-79)^(2/5)=64 I have one solution which is 32847 but apparently there is more than one. Help please!
when you take the 5/2 root of 64, you should end up with both positive and negative root
What do you mean by that
x-79=64^(5/2) =+-32768 so x=32847 or -32689 When you take a square root you end up with +-.
oooooo okay so you have to plug it back in after that makes sense now
I have another question also
yes
Yeah I still don't really understand how to get those answers haha
I mean: \[\huge \color{blue}{(\sqrt{64})^5}\]
@OOOPS , you are interpreting 5/2 as 2/5 there..
too much interference. I cannot provide a cogent answer.
Thanks @waterineyes for assistance.
Can I have all of your help on another question too? I've tried this one countless times and cannot seem to find an answer
\[\huge \sqrt[5] {64^2} \ne (64)^{\frac{5}{2}}\]
The perimeter of a rectangle is 200 feet. Describe the possible lengths of a side if the area of the rectangle is not to exceed 900 feet. ANSWER: The length (in feet) of a side is in the interval ___ or in the interval ____
ok @oops
@OOOPS you are not at all interfering, thing is you are interpreting 5/2 as 2/5. and this way you are marching towards wrong path, Hope you understand what you are interpreting wrong.. @DominicNg is correct in his explanation..
@emmyvirkus please ask as a new question.
If you like me to explain further why there are 2 answers I can.
Okay please do!
\[(x-79)^\frac{ 2 }{ 5 }=64\]so\[(x-79)^\frac{ 1 }{ 5 }=\sqrt{64}=\pm8\]
\[x-79=(\pm8)^5\]
I suppose you are ok with the rest
where did you get the square root for 64 from
oh because 1/5 squared is 2/5?
let's take a very simple example \[x^2=64\]then\[x=\sqrt{64}=\pm8\]ok with that?
Yes you are right
Everything alright now?
so I'm taking the square root of (2/5) and of 64 but how do I know that the square root of 2/5 is 1/5? Because when i put it in my calc it comes out as a funky decimal
or should I just know that by common sense lol
another way of seeing it is\[(x-79)^\frac{ 2 }{ 5 }=[(x-79)^\frac{ 1 }{ 5 }]^2\]
So you square root both sides first
then you raise both sides to the 5th power
I dont understand why you don't just take the reciprical of the exponent from the original equation than just add 79 to each side
is that just not the way to solve this particular kind of problem? because that's how I got my first answer
Yes, that is what I did too. I am explaining why there is a plus and minus sign. This comes about because of the square rooting operation. The square rooting operation is hidden in the 'reciprocal of the power' so my purpose is to highlight it.
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