Solution check for Logarithmic equation

[opus]

Homework Statement

Solve:

$ln(x-4)-ln(x)=6$

I know the solution is no solution, but the value I got is what I am unsure of as I could have got a million different values for x that would result in no solution, but I'm not sure if the value that I got is correct. Once you hit the "no solution" button for the homework, it tells you if you're right or wrong and doesn't tell you the correct value of x.

Homework Equations

The Attempt at a Solution

(i) $$ln\left(\frac{x-4}{x}\right)=6$$

(ii) $$e^{ln\left(\frac{x-4}{x}\right)}=e^6$$

(iii) $$\frac{x-4}{x}=6$$

(iv) $$x-4=6x$$

(v) $$-4=5x$$

(vi) $$x=\frac{-4}{5}$$

(v) x=negative so there is no solution

Step (iii) is where I'm questioning myself. I know that step (iv) results from (iii), but I can't seem to justify (iii) and how I just made the expressions exponents of the base e.

 

[SammyS]

Homework Statement

Solve:

$ln(x-4)-ln(x)=6$

I know the solution is no solution, but the value I got is what I am unsure of as I could have got a million different values for x that would result in no solution, but I'm not sure if the value that I got is correct. Once you hit the "no solution" button for the homework, it tells you if you're right or wrong and doesn't tell you the correct value of x.

You ask a rather strange question here, because if there's no solution, then there is not a correct value for $x$.
The method you used gave an "extraneous" solution, but there are ways to analyze this problem which lead directly to the conclusion that there is no solution without generating an extraneous solution. (See Post #12. below.)

It seems to me that you mean to ask if your work leading up to the extraneous solution is correct. The short answer to that is: No, you have an error.

Homework Equations

The Attempt at a Solution

(i) $$\ln\left(\frac{x-4}{x}\right)=6$$

Although the given equation has no solution, the above equation does have a solution. That solution is not $\displaystyle \ x=\frac{-4}{5}\,.$

Note: Added in Edit.
If you had not made a mistake, your extraneous solution wold have satisfied the equation in (i) .​

(ii) $$e^{ln\left(\frac{x-4}{x}\right)}=e^6$$

Simplifying the above equation does not give the following. Check the right hand side.

(iii) $$\frac{x-4}{x}=6$$

(iv) $$x-4=6x$$
(v) $$-4=5x$$
(vi) $$x=\frac{-4}{5}$$
(v) x=negative so there is no solution

Step (iii) is where I'm questioning myself. I know that step (iv) results from (iii), but I can't seem to justify (iii) and how I just made the expressions exponents of the base e.

 

[opus]

You make a good point! Glad you caught on to that. What I meant to say, and what you correctly assumed, was in asking if my algebraic steps to get to the conclusion that there is no solution were correct. Let me have a look at what you've said in regards to my steps to see where I made my mistake.

 

[opus]

Just to be clear, are you saying that step (ii) is correct up to that point, and something below it is off, or that step (ii) is incorrect? My train of thought for (ii) was to raise both expressions as an exponent of e, that way, they'd have the same base and I could equate the exponents. But saying $e^6$ turns into 6 in the next step doesn't sit right with me, even though I do believe that you can indeed equate the exponents in that case.

 

[Mark44]

What @SammyS was saying was that your step iii has an error.

If $e^{\ln\left(\frac{x - 4} x\right)} = e^6$, then it doesn't follow that $\frac {x - 4} x = 6$.

$e^{\ln(\text{whatever})} = \text{whatever}$ (providing $\text{whatever} > 0$.

 

[Ray Vickson]

Homework Statement

Solve:

$ln(x-4)-ln(x)=6$

I know the solution is no solution, but the value I got is what I am unsure of as I could have got a million different values for x that would result in no solution, but I'm not sure if the value that I got is correct. Once you hit the "no solution" button for the homework, it tells you if you're right or wrong and doesn't tell you the correct value of x.

Homework Equations

The Attempt at a Solution

(i) $$ln\left(\frac{x-4}{x}\right)=6$$

(ii) $$e^{ln\left(\frac{x-4}{x}\right)}=e^6$$

The equation
$$\ln \left( \frac{x-4}{x} \right) = 6$$
becomes
$$ \frac{x-4}{x} = e^6\;\;\Leftarrow\; \text{YES} $$
not what you wrote above, that is, NOT
$$ \frac{x-4}{x} = 6\;\; \Leftarrow \; \text{NO}$$

 

[opus]

Okay, how about this, going from:

(i) $$\frac{x-4}{x}=e^6$$

(ii) $$x-4=xe^6$$

(iii) $$x-xe^6=4$$

(iv) $$x\left(1-e^6\right)=4$$

(v) $$x=\frac{4}{1-e^6}$$

(vi) $$x≈-0.009$$ ⇒ No Solution

 

[SammyS]

Okay, how about this, going from:
(v) $$x=\frac{4}{1-e^6}$$
(vi) $$x≈-0.009$$ ⇒ No Solution

Actually, $\displaystyle \ \frac{4}{1-e^6} = -0.00993964... \,.$

Round that to −0.0099 or −0.010 .

Plug either of these into $\displaystyle \ \ln \left( \frac{x-4}{x} \right) \,.\$ Either should give a result close to 6 .

 

[Mark44]

Actually, $\displaystyle \ \frac{4}{1-e^6} = -0.00993964... \,.$

Round that to −0.0099 or −0.010 .

Plug either of these into $\displaystyle \ \ln \left( \frac{x-4}{x} \right) \,.\$ Either should give a result close to 6 .

But the original equation is $\ln(x - 4) - \ln(x) = 6$, not $\ln\left(\frac{x - 4} x\right) = 6$. From this equation, it's easy to see that we must have x > 4 and x > 0, which means that x > 4. This restriction should have been explicitly stated right from the get-go...

Since the solution found is negative, the equation has no solution.

 

[Ray Vickson]

But the original equation is $\ln(x - 4) - \ln(x) = 6$, not $\ln\left(\frac{x - 4} x\right) = 6$. From this equation, it's easy to see that we must have x > 4 and x > 0, which means that x > 4. This restriction should have been explicitly stated right from the get-go...

Since the solution found is negative, the equation has no solution.

To add to that (and, hopefully, to clarify the issue for the OP):
(1) The equation $\ln(x-4) - \ln x = 6$ needs $x > 4.$
(2) The almost-equivalent equation $\ln ((x-4)/x) = 6$ needs $\{x > 4\} \cup \{ x < 0 \}$
(3) The further-reduced equation $(x-4)/x = e^6$ needs $x \neq 0.$

 

[opus]

Thanks everyone. I think step (ii) of my original post is what was giving me confusion, but I think I realize that it's just changing from logarithmic form to exponential form. Skipping (ii) and going directly from (i) to (iii) makes more sense, but that's not how the text has it written. I don't know why, but in step (ii) I feel uneasy about it as it feels like I'm just taking some expressions and making them an exponent to some base.

 

[Ray Vickson]

Thanks everyone. I think step (ii) of my original post is what was giving me confusion, but I think I realize that it's just changing from logarithmic form to exponential form. Skipping (ii) and going directly from (i) to (iii) makes more sense, but that's not how the text has it written. I don't know why, but in step (ii) I feel uneasy about it as it feels like I'm just taking some expressions and making them an exponent to some base.

Why would you feel uneasy? All you are saying is that if $a=b$ then $e^a = e^b$ (along with the definition of $\ln$, which is that $e^{\ln w} = w$ for any $w>0$).

Anyway, converting and solving for $x$ is not necessary; just use the fact that $x-4 < x$, so $\ln (x-4) < \ln x$. That means that the left-hand-side of your given equation is $< 0$ and hence cannot possibly be equal to 6.

 

[opus]

Why would you feel uneasy? All you are saying is that if $a=b$ then $e^a = e^b$ (along with the definition of $\ln$, which is that $e^{\ln w} = w$ for any $w>0$).
Anyway, converting and solving for $x$ is not necessary; just use the fact that $x-4 < x$, so $\ln (x-4) < \ln x$. That means that the left-hand-side of your given equation is $< 0$ and hence cannot possibly be equal to 6.

That makes sense as when I look at it as it's written, it is just a property of exponents. Understood. I guess I was only looking at the property one way, whereas it can be looked at in both directions.

Ok I like that train of thought, and I can see why the numerator of the argument is less than the denominator of the argument. But why does that mean that $LHS\lt0$?

 

[Mark44]

Ok I like that train of thought, and I can see why the numerator of the argument is less than the denominator of the argument. But why does that mean that

Because $y = \ln x$ is a monotone increasing function.

This means that if a < b, then $\ln a < \ln b$. (The "monotone" descriptor eliminates the possibility of $\ln a$ being equal to $\ln b$ when a < b.)

In your original equation, the left side is $\ln(x - 4) - \ln (x)$. For the reason given above, it must be true that $\ln(x - 4) < \ln(x)$, hence this difference has to be negative.