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# how do I evaluate natural log functions

also, how do I simplify e^[ln(5x-1)]

e^[ln(5x-1)]

First know your logarithm formulas. log(base a)b=x is the same as a^x=b; a^(log(base a)b)=b; lne = 1; lna=log(base e)a

according to the last formula I mentioned, e^[ln(5x-1)] = e^[log(base e)(5x+1)].

according to the second formula I mentioned, e^[log(base e)(5x+1)] = (5x+1).

Hope this helps.

This question is testing your understanding of what a natural logarithm is.

There is a special number called "e".  There is some power that you could raise e to that will give you the result of 5.  As a formula, that would look like this:

ex = 5.

That x is defined as the natural logarithm of 5.  It's the exponent of e that would make e to the x = 5.  Just put ln(5) in place of the x like this:

eln(5)=5

Looking at this equation, you can see that the argument of the natural log (the 5) is the value of e raised to the power ln(5), so:

eln(5x-1)=5x-1

You will use this relationship to do what they call putting a logarithmic quantity into "exponential form".  It's really confusing at first because it's not like solving a normal equation.  It's a transformation into a different form. That transformation will let you do some really powerful stuff later.  eln(5x-1) and (5x-1) are completely interchangeable.  They are in fact the exact same quantity, so you don't have to balance the transformation on the other side of an equation or anywhere else.  It's no worse than rewriting 6 and (4+2).  It looks different, but it may let you something that you couldn't easily do otherwise.

ex and ln x are inverse to each other. So, e[ln(5x-1)] = 5x-1

Attn: f-1(f(x)) = x

Let U be any function whose domain is satisfiable for the natural log function: Over the Reals, Dom(U) > 0.

Then e^[ln(U)] = U.

Likewise ln[e^U] =U.