# Thread: Need Help Solving a Proof for a logarithmic function

1. ## Need Help Solving a Proof for a logarithmic function

Hello, I'm currently stuck and need some help solving this proof.

x^log(Y) = y^log(X)

i am not really sure where to start..

i thinking about using the log definition first, something like

Y = log(X)

X = b^Y

and then taking log of both sides to get

log(X) = log(Y)

though, i am pretty lost from here. I am not even sure if I am doing this right, since I was told to work backwards first, then fowards.

any help / suggestions would be awesome !

2. Originally Posted by jimbutler83
Hello, I'm currently stuck and need some help solving this proof.

x^log_b(Y) = y^log_b(X)

i am not really sure where to start..

i thinking about using the log definition first, something like

...
That's exactly the right idea :

$x = b^{\log_b(x)}$ and $y = b^{\log_b(y)}$

Plug in these terms into the given equation:

$x^{\log_b(y)} = y^{\log_b(x)}~\implies~b^{\log_b(x) \cdot \log_b(y)} = b^{\log_b(y) \cdot \log(x)}$

Can you take it from here?

3. Hello, jimbutler83!

You're on the right track . . .

Prove: . $X^{\log(Y)} \:=\: Y^{\log(X)}$ . The base of the logs is not important.

$\text{Let: }\:X^{\log(Y)} \:=\:P$ .[1]

Take logs: . $\log\left(X^{\log(Y)}\right) \:=\:\log(P) \quad\Rightarrow\quad \log(Y)\!\cdot\!\log(X) \:=\:\log(P)$

We have: . $\log(X)\!\cdot\!\log(Y) \:=\:\log(P) \quad\Rightarrow\quad \log\left(Y^{\log(X)}\right) \:=\:\log(P)$

Hence: . $Y^{\log(X)} \:=\:P$ .[2]

Equate [1] and [2]: . $X^{\log(Y)} \;=\;Y^{\log(X)}$

4. Originally Posted by Soroban
Hello, jimbutler83!

You're on the right track . . .

$\text{Let: }\:X^{\log(Y)} \:=\:P$ .[1]

Take logs: . $\log\left(X^{\log(Y)}\right) \:=\:\log(P) \quad\Rightarrow\quad \log(Y)\!\cdot\!\log(X) \:=\:\log(P)$

We have: . $\log(X)\!\cdot\!\log(Y) \:=\:\log(P) \quad\Rightarrow\quad \log\left(Y^{\log(X)}\right) \:=\:\log(P)$

Hence: . $Y^{\log(X)} \:=\:P$ .[2]

Equate [1] and [2]: . $X^{\log(Y)} \;=\;Y^{\log(X)}$

thanks so much, though i was wondering if it is possible to explain this step a bit more clearly..

why does

$\log(X)\!\cdot\!\log(Y) \:=\:\log\left(Y^{\log(X)}\right)$

is this some kind of log property ?

thanks again.

5. Originally Posted by jimbutler83
thanks so much, though i was wondering if it is possible to explain this step a bit more clearly..

why does

$\log(X)\!\cdot\!\log(Y) \:=\:\log\left(Y^{\log(X)}\right)$

is this some kind of log property ?

thanks again.
Yes.

$a^n~\implies~\log(a^n) = n \cdot \log(a)$

The exponent becomes a factor of the log-value.

If you have a factor at a log-term this factor can be expressed as an exponent. That's what Soroban did.