Did I solve this equation 4*ln(3x)=15 or is it incomplete?

[Witcher]

Homework Statement

I have this problem that seems to be too easy to be true. I don’t know if i am done and would like to move on to the next problem.

Relevant Equations

4Ln3x=15

I stopped where it says e^15=3x^4 it seems complete but i am not sure.

 

[OmCheeto]

Homework Statement:: I have this problem that seems to be too easy to be true. I don’t know if i am done and would like to move on to the next problem.
Homework Equations:: 4Ln3x=15

I stopped where it says e^15=3x^4 it seems complete but i am not sure.

Seems like an odd place to stop. Aren't you trying to solve for x?

 

[SammyS]

Homework Statement:: I have this problem that seems to be too easy to be true. I don’t know if i am done and would like to move on to the next problem.
Homework Equations:: 4Ln3x=15

I stopped where it says e^15=3x^4 it seems complete but i am not sure.

In addition to what @OmCheeto said, you have made an error in applying the rule:

$m \cdot \ln(A) =\ln(A^m)$​

Rather than $\ln(3x^4)$, you should have $\ln((3x)^4)$.

 

[Witcher]

Seems like an odd place to stop. Aren't you trying to solve for x?

Yea just solving for x but it’s X^4. I may have a few more things in mind that may work.

 

[Mark44]

As already noted, your second line is wrong. Instead of $\ln 3x^4 = 15$, you should have what @SammyS wrote in his post.

A different approach would be to divide both sides by 4 as your first step, to get $\ln (3x) = \frac {15} 4$. From there it's only two steps to get the solution.

 

[HallsofIvy]

Are you not clear on what "solve an equation" means?

To solve an equation in x means to find the value or values of x that satisfy the equation. No, writing $3x^4= e^{15}$ has NOT solved the equation (and what you have done is incorrect). You must have an equation of the form "x= something".

Starting from $4ln(3x)= 15$, divide both sides by 4 to get $\ln(3x)= \frac{15}{4}$. Then take take the exponential of both sides $3x= e^{\frac{15}{4}}$. Last, divide both sides by 3: $x= \frac{1}{3}e^{\frac{15}{4}}$.

Another way to approach this problem, and more like what you probably did, is, instead of dividing both sides of the equation, take the "4" inside the logarithm: $$ln((3x)^4)= 15$$. Notice that it is both "3" and "x" that are to the 4th power. This is NOT "$3x^4$" it is $(3x)^4= 81x^4$. So $ln(81x^4)= 15$ and taking the exponential of both sides $81x^4= e^{15}$. $x^4= \frac{e^{15}}{81}$ and then take the fourth root of both sides: $x= \frac{(e^{15})^{1/4}}{\sqrt[4]{81}}= \frac{e^{15/4}}{3}$ as before.

 

[symbolipoint]

My crude but practical steps are show here but in pure text.

ln(3x)=15/4

e^(15/4)=3x

x=(e^(15/4))/3

 

[Ojile Joseph]

With the approaches above, you are sure covered. Maths& science is fun.