# Math Help - Natural Logarithums

1. ## Natural Logarithums

HI....Sorry to bother you all. I posted on here before with a problem I have been having. As I am new here I am getting to grips with using the great site and getting some help from people in the know. Please find the thumb nail attached. I thought I knew what I was doing but have really lost me way. If some one could lay this out so I could understand that would be very helpful.

2. ## Re: Natural Logarithums

Originally Posted by camorris
HI....Sorry to bother you all. I posted on here before with a problem I have been having. As I am new here I am getting to grips with using the great site and getting some help from people in the know. Please find the thumb nail attached. I thought I knew what I was doing but have really lost me way. If some one could lay this out so I could understand that would be very helpful.
$L = {L_0}\left( {{e^{h\omega }} + 1} \right)$

3. ## Re: Natural Logarithums

So the problem is: given $\omega= \frac{1}{h} ln\left(\frac{L}{L_0}-1\right)$, find L when $\omega= -2.6$, $L_0= 16$ and $h= 1.5$.

Putting in the given values, $-2.6= \frac{1}{1.5}ln\left(\frac{L}{16}- 1\right)$.

Notice what that says- if you divide L by 16, then subtract 1, take the natural logarithm, and divide by 1.5, then the result is -2.6. To go the opposite way, from -2.6 to a value of L, you have to do the opposite- do the opposite of each operation, in the opposite order. Since the last thing done above was "divide by 1.5", the first thing we should do to solve for L is "multiply by 1.5". Doing that on both sides gives $-2.6(1.5)= -3.9= ln\left(\frac{L}{16}- 1\right)$. The opposite of "natural logarithm", ln, is its "inverse function" the exponential. Taking the exponential of both sides gives $e^{-3.9}= 0.0202= \frac{L}{16}- 1$. The opposite of "subtract 1", -1, is "add 1". Adding 1 to both sides, $0.0202+ 1= 1.0202= \frac{L}{16}$. Finally, the opposite of "divide by 16" is, of course, "multiply by 16". Multiplying both sides by 16, $(0.0202)(16)= 0.3239= L$

4. ## Re: Natural Logarithums

Thank you so much. I did it the first time and got the same format as Plato has provided although I did the (e^hw - 1).

I see I have to formats to go from and to different ways. Thank you very much once again.