Trouble with logarithms?

Question

i don't understand how to simplify a lot of stuff like

e^ln3

e^3lnx

e^(1+lnx)

please i dont want just the answers i want explanations

Answer 1

Here is the trick:

e^(ln[x]) = x

Thats right, the exponetial fuction cancels the natural logarithm and vice versa, i suppose you could say that they are the opposite of each other?

Anyway for example

e^(ln[3]) = 3

Now the second one involves the previous rule as well as this one:

e^(2a) = e^[a]^2

So..

e^(3ln[x])

= e^(ln[x])^3

= x^3

Now the third one involves the first rule as well as this one:

e^(a) * e^(b) = e^(a+b)

Therefore:

e^(1 + ln[x])

= e^(1) * e^(ln[x])

= e^(1) * x

=xe^(1)

This last one gives a definite answer of around 27

Hope this helps!

Answer 2

y = e^(ln3)

ln y = (ln 3) ln e

ln y = ln 3

y = 3

e^(ln 3) = 3

y = e^(3 ln x)

ln y = (3 ln x) ln e

ln y = 3 ln x

ln y = ln x³

y = x³

e^(3 ln x) = x³

( e ) e^(ln x)

e x

Answer 3

e^ neutralizes ln so e^ln(3) = 3 (take whatever in the parantheses)

e^3lnx can be rewritten to e^ln(x^3) - this is a logarithm rule (see link) after rewriting you can do the same as number one. answer is x^3 (the number between the parantheses)

e^(1+lnx) can be rewritten to e^1 . e^lnx (logarithm rule) = e^1 . x  (. = multiplied by)

Answer 4

e ^ ln 3 =e ^ 1.1098612289=3

the other two you must know the value of x