Question

Can you please help guide me through the last steps in this equation involving logarithms?

Answer 1

\[1+\ln x = \ln4\]

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

The properties I used were


ln(x) - ln(y) = ln(x/y)


and


ln(x) = y converts to e^y = x

Answer 6

I see the division coming into play with the subtraction.

Answer 7

The last one is the key I was missing.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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