Here's the question you clicked on:

## FreeTrader Group Title Can you please help guide me through the last steps in this equation involving logarithms? 2 years ago 2 years ago

• This Question is Closed
1. FreeTrader Group Title

$1+\ln x = \ln4$

2. FreeTrader Group Title

I am solving for x. Can you nudge me about the ln properties?

3. jim_thompson5910 Group Title

1+ln(x) = ln(4) 1 = ln(4)-ln(x) 1 = ln(4/x) 4/x = e^1 4/x = e 4 = ex 4/e = x x = 4/e

4. FreeTrader Group Title

Thanks, I'll see if I can follow this on paper here... Just a sec...

5. jim_thompson5910 Group Title

The properties I used were ln(x) - ln(y) = ln(x/y) and ln(x) = y converts to e^y = x

6. FreeTrader Group Title

I see the division coming into play with the subtraction.

7. FreeTrader Group Title

The last one is the key I was missing.

8. satellite73 Group Title

you can also try $\ln(x)=\ln(4)-1$ $x=e^{\ln(4)-1}=e^{\ln(4)}\times e^{-1}=\frac{4}{e}$

9. FreeTrader Group Title

ln x = e to what power gives us x Logarithms confuse me to no end, despite how "simple" they are supposed to be.

10. FreeTrader Group Title

I have duplicated the algebra, and it makes sense. Are you aware of a really good ln tutorial with exercises?

11. jim_thompson5910 Group Title

try this page: http://www.purplemath.com/modules/solvelog.htm

12. FreeTrader Group Title

Thank you. If you answer that tutorial question in the new Q I posted, I'd be pleased to pin a medal upon you.