What is the solution of the equation?
\[4(3-x)^{\frac{ 4 }{ 3 }-5}=59\]
Alright. Do you want to try it on your own first, and see if you can get an answer?
Sure, I'll write it down on paper because solving math on here isn't the easiest.
I agree. You go ahead and try it, if you cannot get it, we will walk through step by step :) No rush
Maybe I will start writing the steps, because I might have to leave and get pizza in a few. xD
What we are going to do is take 4(3-x)^4/3 - 5 = 59 . Lets add that (-5) to both sides to get; 4(3-x)^4/3 = 64
From here, we will try to remove the "4" so we will have -- > (3-x)^4/3 instead. So lets divide 4 from each side. We will then have (3-x)^4/3 = 16
Now we use the principle of fraction exponents to radicals
A^B/C = c(sqrt(a))^B
\[\sqrt[3]{3-x}^{4} = 16\]
We use the principle of even roots, in this case 4th roots. When taking even roots we must use ± on the right side. Now since 16 = 2·2·2·2 = 24, then the 4th root of 16 is 2. So we have:
\[\sqrt[3]{3-x} = \pm 2\]
Next to get rid of the cube root we cube both sides........
(Simply distribute it)
3-x = ± 8
If we use the " + " we get; 3-x = 8 -x = 5 x = -5
If we use the - we get 3-x = -8 -x = -11 x = 11
Both are valid solutions, when put into the original equation.
Just put them where 'x' is to check, if you are unsure.
Sorry if it was messy or rushed, I had to kind of hurry. xD I wanted to make sure you had the answer in time.
No, that's fine! That makes sense now. Question, when you cubed both sides, why did you drop the exponent of 4?
Because of this...: We use the principle of even roots, in this case 4th roots. When taking even roots we must use ± on the right side. Now since 16 = 2·2·2·2 = 24, then the 4th root of 16 is 2. So we have:\[\sqrt[3]{3-x} = \pm 2\] By taking the 16 and doing .... 2 x 2 x 2 x 2 = 2^{4} this sort of cancels it out, I guess
Okay, makes sense. Thank you so much! :)):):):):):):)
Not a problem, if you have any more questions, please feel free to tag me!
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