strawberrylatent:

If f(x)=1/(3x-2), then f^-1(x)=?

Also

Simplify y^(2/7) * y^(14/2)

And one more

Simplify (y^(2/7))^(14/2)

f(x) = 1/(3x-2)

To find f^-1(x), you switch x and f(x). For ease of typing, lets use y instead of f(x). Then, y=1/(3x-2). Switching, you get x=1/(3y-2). Then you solve for y. Multiply both sides by (3y-2), and you have 3xy-2x=1. Then adding 2x to each side is 3xy=2x+1. Finally, divide each side by 3x, and you have y=2/3*1/3x=f^-1(x)

y^(2/7)*y^(14/2)

When you’re multiplying two things with the same base, you simply add the powers. Thus,

y^(2/7)*y^(14/2)=y^(2/7+14/2)

Then you need a common denominator to add the fractions. The least common multiple of 7 and 2 is 14 (which is 7*2), you you can then get

y^(4/14+88/14)=y^(92/14)=y^(46/7)

(y^(2/7))^(14/2)

When simplifying expressions like this, you need to multiply the two exponents together. Thus, (2/7)*(14/2)=(2*14/7*2). Note the 2’s cancel, leaving 14/7, which simplifies to 2. Then, you have

(y^(2/7))^(14/2)=y^2

I hope that helped! The answers should be right. Although I didn’t double check my work… The methods are definitely correct, though!