# ln e^3 + ln e^3x = 6

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ln e^3 + ln e^3x = 6

We know that:

ln e^a = a*ln e = a*1 = a

Then:

ln e^3 + ln e^3x = 3 + 3x = 6

Now subtract 3 :

==> 3x = 3

Divide by 3:

**==> x= 1**

To solve lne^3 +lne^3x = 6.

We know by definition of logathm that

lne^3 = 3 and ln e^3x = 3x.

So the equation could be rewritten as:

3+3x = 6.

3x = 6-3 = 3

3x/3 = 3/3 = 1

x= 1.

We use the relations of logarithms that ln a^b = b * ln a,

and ln a + ln b = ln a*b

ln e^3 + ln e^3x =6

=> ln ( e^3* e^3x) = 6

=> ln [e^ (3+3x)] = 6

=> ln e^( 3+3x) =6

Now ln e^a = a

=> ln e^( 3+3x) =6

=> 3+3x = 6

=> 3x = 3

=> x= 1

**Therefore x = 1.**

Let's try the following method. We'll subtract ln e^3 both sides:

ln e^3x = 6 - ln e^3

We'll re-write 6 as:

6 = 6*1

We'll substitute the value 1 by ln e.

6 = 6*ln e

We'll use the power property of the logarithms:

6 = ln e^6

We'll re-write the equation:

ln e^3x = ln e^6 - ln e^3

Because the bases are matching, we'll transform the difference of logarithms from the right side, into a quotient:

ln e^3x = ln (e^6/e^3)

Because the bases are matching, we'll apply the one to one property:

e^3x = e^(6-3)

e^3x = e^3

Because the bases are matching, we'll apply the one to one property:

3x = 3

We'll divide by 3 both sides:

**x = 1**

The logarithmic equation ln e^3 + ln e^3x = 6 has to be solved.

Use the property of logarithm log a + log b = log a*b. This gives:

ln (e^3*e^(3x)) = 6

The logarithm of a number x with respect to base b is defined such that if `log_b x = y` , `x = b^y` . As ln refers to the natural logarithm, which is logarithm to base e, this gives:

(e^3*e^(3x)) = e^6

e^(3 + 3x) = e^6

3 + 3x = 6

3x = 3

x = 1

The solution of the equation ln e^3 + ln e^3x = 6 is x = 1.