# anti-logging

• May 3rd 2010, 06:31 AM
stealthmaths
anti-logging
Hi:
If I have an equation of the sort

$\displaystyle \ln \left( {3y - 1} \right) = \frac{x}{2}$

and I want to solve for y, I can anti-log which gives:

$\displaystyle 3y - 1 = e^{\frac{x}{2}}$

The problem is, I can't understand how the x/2 has now become a power. Because if I multiply ln by e to give lne=1, and I do the same for the RHS which gives ex/2, not $\displaystyle e^{\frac{x}{2}}$

can someone help me to understand this please
thank you
• May 3rd 2010, 06:41 AM
stealthmaths
ahhh!
I may be able to answer my own question here.
Is it because

$\displaystyle \ln x = y{\rm{ }} \Leftrightarrow {\rm{ e}}^y = x$?

So I shouldn't be thinking about algebraically multiplying both sides by e, but instead be applying the log laws?
• May 3rd 2010, 10:13 PM
CaptainBlack
Quote:

Originally Posted by stealthmaths
Hi:
If I have an equation of the sort

$\displaystyle \ln \left( {3y - 1} \right) = \frac{x}{2}$

and I want to solve for y, I can anti-log which gives:

$\displaystyle 3y - 1 = e^{\frac{x}{2}}$

The problem is, I can't understand how the x/2 has now become a power. Because if I multiply ln by e to give lne=1, and I do the same for the RHS which gives ex/2, not $\displaystyle e^{\frac{x}{2}}$

There is no multiplication involved, just the definition of a logarithm.

In other language $\displaystyle \ln(x)$ is not a number like thing by which $\displaystyle x$ is multiplied but an operator which transforms its input ($\displaystyle x$) in to its output as specified by the definition of the logarithm.

CB

CB
• May 4th 2010, 03:32 AM
stealthmaths
ahhh. I see.
That's very clear. Thank you Captain.(Bow)