2(3y-4)^(3/5)-4=50
2(3y-4)^(3/5) = 54
(3y-4)^(3/5) = 27
3y - 4 = 27^(5/3)
3y = 27^(5/3) + 4
y = (27^(5/3) + 4)/3
= (243 + 4)/3 = 247/3 or 82 1/3
1-(1/a)/1-(1/a^2)
multiply top and bottom by a^2 to get
(a^2-a)/(a^2-1)
= a(a-1)/(a+1)(a-1)
= a/(a+1)
or, since you seem to like reciprocals so much,
1 /(1+1/a)
:-)
To solve the equation 2(3y-4)^(3/5) - 4 = 50, you can follow these steps:
Step 1: Add 4 to both sides of the equation to isolate the term with the exponent:
2(3y-4)^(3/5) = 54
Step 2: Divide both sides of the equation by 2 to isolate the term with the exponent:
(3y-4)^(3/5) = 27
Step 3: Raise both sides of the equation to the power of 5/3 to remove the fractional exponent:
[(3y-4)^(3/5)]^(5/3) = 27^(5/3)
Simplifying the left side of the equation:
(3y-4)^(3/5 * 5/3) = 27^(5/3)
(3y-4) = 27^(5/3)
Step 4: Now we need to find the cube root of 27 and raise it to the power of 5 to get our answer:
27^(5/3) ≈ 243^(1/3) ≈ 3^5 = 243
So the equation becomes:
3y - 4 = 243
Step 5: Now, add 4 to both sides of the equation to isolate the variable:
3y = 243 + 4
3y = 247
Step 6: Finally, divide both sides of the equation by 3 to solve for y:
y = 247/3
So the solution to the equation 2(3y-4)^(3/5) - 4 = 50 is y = 247/3.