Log and Solve for X Help

[mowgli8798]

I have been working on this one for a while and wanted to get some other people's opinions.
I was told the answer is x = 1/pi*e but I cannot prove it.

This is what they gave me to start with.

log(pi)*log(pi*x) = log(e)*log(e*x)

without using decimals or derivatives.

Any feed back, advice, or help is much appreciated! Thanks!

 

[Deleted member 4993]

I have been working on this one for a while and wanted to get some other people's opinions.
I was told the answer is x = 1/pi*e but I cannot prove it.

This is what they gave me to start with.

log(pi)*log(pi*x) = log(e)*log(e*x)

Any feed back, advice, or help is much appreciated! Thanks!

Do have log₁₀ or log_e?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[topsquark]

That would be [math]x = \dfrac{1}{e ~ \pi }[/math] or x = 1/(e Pi). You need the parenthesis!

I'm assuming this is log = ln...

[math]ln( \pi ) ~ ln( \pi x ) = ln(e) ~ ln( e x)[/math]
[math]ln( \pi ) ~ ( ln( \pi ) + ln(x) ) = ln(e) ~ (ln(e) + ln(x) )[/math]
[math]ln( \pi )~ ( ln( \pi ) + ln(x) ) = 1 \cdot (1 + ln(x) )[/math]
Now solve for ln(x) and take the exponential of it. It's not terribly hard but there's one simplification near the end that you have to spot out. Do it one little step after another.

-Dan

 

[mowgli8798]

It's log10 not loge. I have tried two things mainly. The first is to put all terms on one side, subtracting log(e)*log(e*x). My second attempts, I split log(e*x) and log(pi*x) into log(e) + log(x) and log(pi) + log(x) respectively. Than divided by log(e). The main problem I am having is I do not know how to get rid of the logs and get into the variables and the constants. Because per the reasoning of the problem I am not allowed to use decimals, making simplifying out right like I have in the past not an option. Hope this helps. Let me know if you have any advice.

 

[mowgli8798]

That would be [math]x = \dfrac{1}{e ~ \pi }[/math] or x = 1/(e Pi). You need the parenthesis!

I'm assuming this is log = ln...

[math]ln( \pi ) ~ ln( \pi x ) = ln(e) ~ ln( e x)[/math]
[math]ln( \pi ) ~ ( ln( \pi ) + ln(x) ) = ln(e) ~ (ln(e) + ln(x) )[/math]
[math]ln( \pi )~ ( ln( \pi ) + ln(x) ) = 1 \cdot (1 + ln(x) )[/math]
Now solve for ln(x) and take the exponential of it. It's not terribly hard but there's one simplification near the end that you have to spot out. Do it one little step after another.

-Dan

I can use natural logs like you did. I am not sure how to solve after you got to your last step. That is exactly where I always get hung up. I could subtract one from the right side to get ln(x) alone on the right side. After that I do not know how to get rid of the natural log.

 

[Deleted member 4993]

I can use natural logs like you did. I am not sure how to solve after you got to your last step. That is exactly where I always get hung up. I could subtract one from the right side to get ln(x) alone on the right side. After that I do not know how to get rid of the natural log.

Why did you NOT mention this work in your first post? If you had, you would have been lot nearer to the solution!!!

Next step is to distribute both sides and factor out ln(x).

Please share your work.

 

[mowgli8798]

Okay. After distributing I have ln(pi)^2+ln(pi)*ln(x) = 1 + ln(x). I am not sure how I would factor out x here with the natural logs. Very new to working with logarithms.

 

[mowgli8798]

Okay. After distributing I have ln(pi)^2+ln(pi)*ln(x) = 1 + ln(x). I am not sure how I would factor out x here with the natural logs. Very new to working with logarithms.

if I factor x out of the left side I get [MATH]ln(x)(ln(pi)^2/ln(x)+ln(pi))[/MATH]I divide both sides by x and
I get stuck here: [MATH]ln(pi)^2 / ln(x) + ln(pi) = 1 / ln(x) + 1[/MATH]

 

[Deleted member 4993]

Okay. After distributing I have ln(pi)^2+ln(pi)*ln(x) = 1 + ln(x). I am not sure how I would factor out x here with the natural logs. Very new to working with logarithms.

[ln(pi)]^2+ln(pi)*ln(x) = 1 + ln(x)

ln(π} * ln(x) - ln(x) = 1 - [ln(π)]^2

ln(x) * [ln(π} - 1] = 1 - [ln(π)]^2...................edited...extra * removed

remember

a^2 - b^2 = (a + b) * (a - b)

continue

 

[mowgli8798]

Ok I have proceeded a bit farther. And wanted to double check that I am doing the right operations.

I divided both sides by -ln(pi), and got [MATH]ln(x) = 1 / ln(pi) + ln(pi) [/MATH]
I can tell that I am getting a lot closer. But I again ran into the situation of where I do not know how to resolve the ln. I need to turn ln(x) into x without a bunch of decimals.

 

[Cubist]

ln(π} * ln(x) - ln(x) = 1 - [ln(π)]^2

ln(x) * [ln(π} * - 1] = 1 - [ln(π)]^2

There's a typo on the second line above, the second * is incorrect, so the line should read

ln(x) * [ln(π} - 1] = 1 - [ln(π)]^2

AND @mowgli8798 here's an extra hint that might enable you to use @Subhotosh Khan 's advice right at the end of post#9

ln(x) * [ln(π} - 1] = 1^2 - [ln(π)]^2

 

[mowgli8798]

In this case a = ln(pi) and b = 1? I am still working through this one, but I am having a hard time applying a^2 - b^2

 

[Steven G]

Whenever you have one thing squared - another thing squared you always get (one thing - another thing)(one thing + another thing)
You have 1² - [ln(π)]² so subtract the terms being squared and multiply that by the sum of the terms being squared. Do not let what the terms are bother you. Subtract then and add them!