T^3 = 27 * R^5 is simplified to T=a*R^b

a. Calculate the exact values of a and b

b. If R gets eight times bigger, how many times does T get bigger?

c. The equation for T is simplified to R=p*T^q. Calculate p to one decimal place and q exact.

Guest Feb 20, 2020

#1**+2 **

T^3 = 27 * R^5 is simplified to T=a*R^b

a. Calculate the exact values of a and b

b. If R gets eight times bigger, how many times does T get bigger?

c. The equation for T is simplified to R=p*T^q. Calculate p to one decimal place and q exact.

a. T^3 = 27 * R^5 take the cube root of both sides

T = 27^(1/3) * (R^5)^(1/3)

T = [ 3 R^(5/3) ] so...... a = 3 and b =(5/3)

b. 3 * (8R)^(5/3) =

3 * (8)^(5/3) * (r)^(5/3) =

3 * ( 8 ^(1/3))^5 * R^(5/3) =

3* (2)^5 * R^(5/3) =

32 * [ 3 * R^(5/3) ]

So....T will increase by 32 times when R increases by 8 times

c. Sorry.....I don't know the answer to this one.....

CPhill Feb 20, 2020

#3**+1 **

\(R=p*T^q\\ \)

\(T=3R^{(5/3)}\\ \frac{T}{3}=R^{(5/3)}\\ (\frac{T}{3})^{(3/5)}=R\\ R=(\frac{T}{3})^{(3/5)}\\ R=(\frac{1}{3})^{(3/5)}*T^{(3/5)} \\~\\ p=(\frac{1}{3})^{(3/5)}\\ p=3^{(-0.6)}\\ p\approx 0.517\\~\\ p\approx 0.5\qquad(1\;dec\;place)\\ q=3/5=0.6\qquad (exact) \)

Melody Feb 20, 2020

#4**+1 **

Hi thanks for the reply! This might sound stupid but I was able to follow it all until the p=(1/3)^(3/5). How do you get from this to p=3(-0,6)?

Guest Feb 20, 2020

#6**+1 **

It is never stupid to ask a genuine question. However, if you do not understand then it would be stupid to not ask.

\(x^{-y}=\frac{1}{x^y}\\ p^{-2}=\frac{1}{p^2}\\ etc\\ \text{that is what a negative power means}\\ so\\ (\frac{1}{3})^{3/5}\\ =(\frac{1}{3})^{0.6}\\ =\frac{1}{3^{0.6}}\\ =3^{-0.6}\)

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Think about how index laws work

\(x^3\div x^7=x^{3-7}=x^{-4}\\ but\\ x^3\div x^7=\frac{x^3}{x^7}=\frac{xxx}{xxxxxxx}=\frac{1}{x^4}\\ therefore\\ x^{-4}=\frac{1}{x^4}\\\)

By the way, your post was blanked out because you used the word 'stupid'

It was blanked out for me too but I have the power to unblank

Melody
Feb 20, 2020