I have another math problem that I'm having trouble with, Help please?

Question

Please help me solve:

t 2/3 = 9

I made a mistake with the math problem I had previously posted. The equation was supposed to be written as:

t^2/3 = 9

Answer 1

t = 27/2

Answer 2

As t^2/3=3^2 (as 9=3*3), we can multiply powers of both sides by 3/2, the equation will remain balanced
therefore, t^2/3*3/2=3^2*3/2, we get t=3^3 or 27

Answer 3

t 2/3 = 9 next I will multiply both sides by 3/2 the inverse of 2/3

(3/2)(2/3) t = 9 (3/2)

t = 27/2

t = 13.5

Answer 4

t^(2/3) = 9
( t ^ (2/3) ) ^ (3/2) = 9 ^ (3 / 2)
t = 3 ³
t = 27

Answer 5

you bring over the 2/3 to the other side and change sign so the three comes over to multiple and the 2 comes over to divide

so its

t2/3 = 9
t = 9 x 3/2
t = 27 / 2
t = 13.5

Answer 6

t ^ (2/3) = 9 solve for t
take log of both sides
log(t^(2/3)) = log(9)
using rule log(a^b) = b*log(a)
(2/3) * log(t) = log(9)
log(t) = log(9) / (2/3)
raise both sides to power of 10
10^log(t) = 10^ (log(9) / (2/3))
using rule 10^log(a) = a
t = 10^ (log(9) / (2/3))
solving right hand side
t = 27

Update: Having now seen defeNder's solution method, this is a better way than mine.

Answer 7

t^2/3 = 9
or, (t^1/3)^2 = 3^2
or, t^1/3 = 3
or, ( t^1/3 )^3 = 3^3 [taking cube in both sides]
or, t = 27

so, t = 27 is the answer.

Answer 8

t(2/3)=9

first multiply both sides of the equation with 3

6t=27

second divide both sides with 6

t=27/6

simply factor (3/3)from27/6=9/2

t=9/2

divide 9 with 2

t=4.5

Answer 9

do you mean (2/3) t = 9 or 2t/3 = 9 ???

Answer 10

Given any factor P= (x,y), its distance from the line Ax+ by way of= 0 is given by way of |Ax+ by way of|/sqrt(A^2+ B^2}. Your first line has A= 4, B= 3 so the gap from (x,y) to it extremely is |4x+ 3y|/5. Your 2d line has A= 4, B=-3 so the gap from (x,y) to it extremely is |4x- 3y|/5. announcing "the made of the distances is 4" means that (|4x-3y|/5)(|4x+3y|/5)= 4 or |16x^2- 9y^2|= 20.