Please solve with work and explain why and what you are doing

Write the Inverse of each function:

1) Y= log3X

2) Y= 2^(x/3)

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- Apr 2nd 2008, 05:36 PMnorth1224Logarithms as Inverses.
Please solve with work and explain why and what you are doing

Write the Inverse of each function:

1) Y= log3X

2) Y= 2^(x/3) - Apr 2nd 2008, 05:39 PMJhevon
- Apr 2nd 2008, 05:42 PMTheEmptySet

$\displaystyle y=log_{3}(x)$

exchange the x and y and then solve for y

$\displaystyle x=log_{3}(y) \iff 3^x=y$

$\displaystyle f^{-1}(x)=3^x$

$\displaystyle y=2^{x/3}$ swapping the var

$\displaystyle x=2^{y/3} \iff log_{2}(x)=\frac{y}{3} \iff 3log_{2}(x)=y$

$\displaystyle f^{-1}(x)=3log_2(x)$ - Apr 2nd 2008, 05:54 PMnorth1224Quote:

$\displaystyle

y=2^{x/3}

$ swapping the var

$\displaystyle

x=2^{y/3} \iff log_{2}(x)=\frac{y}{3} \iff 3log_{2}(x)=y

$

- Apr 2nd 2008, 06:01 PMJhevon
- Apr 2nd 2008, 06:11 PMnorth1224
- Apr 2nd 2008, 06:18 PMTheEmptySet
It is the rule that Jhevon mentioned above.

- Apr 2nd 2008, 06:21 PMnorth1224
- Apr 2nd 2008, 06:30 PMTheEmptySet
I don't understand 2 is a constant not a variable?

your equation is

$\displaystyle y=2^{\frac{x}{3}}$

The 2 is the base of eponential function.

if we interchange the variables we get

$\displaystyle x=2^{\frac{y}{3}}$ rewritting as a log we get

$\displaystyle log_{2}(x)=\frac{y}{3} \iff y= 3 log_{2}(x)$ - Apr 2nd 2008, 06:34 PMnorth1224My Bad