algebra II help please!!!?

Question

IN 3x = 6



and


In x = -2




please help me out! and if u can;t, then don't leave rude comments.

Answer 1

ln[3x] = ln[3] + ln[x] = 6
ln[x] = 6 - ln[3]
Raise both sides to e
e^(ln[x]) = x = e^(6-ln[3]) = (e^6)e^(-ln[3]) = (e^6)/3
x = (e^6)/3

ln[x] = -2
e^(ln[x]) = e^(-2)
x = e^(-2)

Answer 2

In order to do these question, you have to know how to convert from exponential form to logarithmic form and vice versa.

The following is expressed in exponential form.

y = a^x

To convert this to logarithmic form, do the following.
(1) The base of the exponent becomes the base of the log.

log[a](?) = ?

(2) The exponent because the answer.

log[a](?) = x

At this point, you just put what's left, inside the log. In this case, it's y.

log[a](y) = x
Is the logarithmic form of
x^a = y

The most important thing to know, whether going forwards or backwards, is to realize that **THE BASE OF THE LOG BECOMES THE BASE OF THE EXPONENT**.

Let's try your questions.

ln(3x) = 6

First, remember that ln is defined to be a log with a base of e. Changing that to exponential form, we get

e^6 = 3x

Therefore, x = (e^6)/3

I want you to solve #2 on your own.

Answer 3

What do you need to find here? Do you know how to convert a logarithm back to normal form? Do you know what ln means? If you answered yes to all the questions and you're looking for x, then this is really very simple to do.

Answer 4

ln 3x = 6 => 3x = e^6 => x = (e^6)/3
ln x = -2 => x = e^-2

Answer 5

i think i need more info
3(-2)=-6