# Logarithms as Inverses.

• Apr 2nd 2008, 05:36 PM
north1224
Logarithms 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 PM
Jhevon
Quote:

Originally Posted by north1224
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)

note that $\displaystyle \log_a b = c \Longleftrightarrow a^c = b$
• Apr 2nd 2008, 05:42 PM
TheEmptySet
Quote:

Originally Posted by north1224
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)

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

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 PM
north1224
Quote:

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

$\displaystyle x=2^{y/3} \iff log_{2}(x)=\frac{y}{3} \iff 3log_{2}(x)=y$
Where did the 2 come from?
• Apr 2nd 2008, 06:01 PM
Jhevon
Quote:

Originally Posted by north1224
Where did the 2 come from?

the 2 was in the original question. it ended up as the base of the logarithm based on the rule i gave you earlier
• Apr 2nd 2008, 06:11 PM
north1224
Quote:

Originally Posted by Jhevon
the 2 was in the original question. it ended up as the base of the logarithm based on the rule i gave you earlier

I Meant the two from "TheEmptySet" description, Thanks anyway, The rule helped me alot for the rest of my worksheet though, thanks for that
• Apr 2nd 2008, 06:18 PM
TheEmptySet
It is the rule that Jhevon mentioned above.
• Apr 2nd 2008, 06:21 PM
north1224
Quote:

Originally Posted by TheEmptySet
It is the rule that Jhevon mentioned above.

No, TheEmptySet, Where did the 2 as a variable come from.?
• Apr 2nd 2008, 06:30 PM
TheEmptySet
Quote:

Originally Posted by north1224
No, TheEmptySet, Where did the 2 as a variable come from.?

I don't understand 2 is a constant not a variable?

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

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 PM
north1224
Quote:

Originally Posted by TheEmptySet
I don't understand 2 is a constant not a variable?

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