1. ## Logarthimic proof

i have the following:

b^h = e^ah

i have been told to take the log of each side and show that there is a value of a for every h. Can i do the following?:

ln b^h = ah

but ln b^h = h ln b

so lnb^h = h ln b = ah

cancel h

ln b^h= lb b = a

divide through by ln b

h = a

???? doesnt read right to me.......

2. Originally Posted by thermalwarrior
i have the following:

b^h = e^ah

i have been told to take the log of each side and show that there is a value of a for every h. Can i do the following?:

ln b^h = ah

but ln b^h = h ln b

so lnb^h = h ln b = ah

cancel h

ln b^h= lb b = a

divide through by ln b

h = a

???? doesnt read right to me.......
Something is screwy:
$\displaystyle b^h = e^{ah}$

$\displaystyle ln(b^h) = ln(e^{ah})$

$\displaystyle h \cdot ln(b) = ah$

$\displaystyle ln(b) = a$

So a has nothing to do with h...

-Dan

3. so is h could just be any value just attached onto the final form?

4. Originally Posted by thermalwarrior
so is h could just be any value just attached onto the final form?
That's correct. You even don't need any logarithms:

$\displaystyle b^h = e^{ah} = \left(e^a \right)^h ~\implies~ b = e^a$