Help With Log Problem

[Ted_Grendy]

Hi All

I was wondering if someone could confirm if the following is correct:

I need to make Y the subject.

Ln(A)+n^2+1 = Ln(Y)

1) Because Ln(e) = 1
2) n^2 = (n^2)Ln(e) = Ln(e)^(n^2)
3) 1 = Ln(e)
4) Ln(A) + Ln(e)^(n^2) + Ln(e) = Ln(A * (e)^(n^2) * (e))
5) Ln(A * (e)^(n^2) * (e)) = Ln(Y)
6) A * (e)^(n^2) * (e) = Y

Thanks

 

[Otis]

Y = A ∙ e^(n^2) ∙ e

I didn't check all your steps, but I see the result is correct. However, it can be simplified.

Remember: m^a ∙ m^b = m^(a + b)



Here's another approach, to solving the equation. Get the logarithmic terms combined into a single logarithm (on one side) and everything else on the other side. Then exponentiate each side. With this approach, you'll end up with the simplified result. :cool:



PS: When the base in a power is a single symbol, you don't need to type grouping symbols around it.

For example, typing e^(n^2) instead of (e)^(n^2) is okay.

 

[Ted_Grendy]

Another Log Problem

Thanks Otis

Any chance I can trouble you for with another log problem?

I need to fin Y in terms of x for the following:-

-3logy+2log(x^2)-logx=logy^(1/3)

I get:-

y = x^3

 

[Otis]

Sure, but in the future, please start a new thread, to discuss a different exercise.



You've used function notation with only one term, not the others. Here's how I'd type the given equation:

-3∙log(y) + 2∙log(x^2) - log(x) = log(y^(1/3))



There are two Real solutions for y. Your solution is close to one of them, but the exponent is incorrect. Please show your work. :cool:



PS: I've guessed that your term logy^(1/3) means log(y^(1/3)) because the Real solution using log(y)^(1/3) instead looks too messy.

 

[JeffM]

Thanks Otis

Any chance I can trouble you for with another log problem?

I need to fin Y in terms of x for the following:-

-3logy+2log(x^2)-logx=logy^(1/3)

I get:-

y = x^3

It truly helps if you show your work so there we can see exactly where you erred. Here is a step by step. I can guess where you made a mistake, but I cannot be sure.

\(\displaystyle -\ 3 log(y) + 2log(x^2) - log(x) = \log(y^{1/3}) \implies\)

\(\displaystyle 2 log(x^2) - log(x) = 3log(y) + log(y^{1/3}) \implies\)

\(\displaystyle log(\{x^2\}^2) - log(x) = log(y^3) + log(y^{1/3}) \implies\)

\(\displaystyle log(x^4) - log(x) = log(y^3) + log(y^{1/3}) \implies\)

\(\displaystyle log(x^4 \div x) = log(y^3 * y^{1/3}) \implies\)

\(\displaystyle log(x^3) = log(y^{10/3}) \text { because } 3 + \dfrac{1}{3} = \dfrac{9}{3} + \dfrac{1}{3} = \dfrac{10}{3}.\)

I suspect that you multiplied 3 and 1/3 and got 1. When you multiply exponentiated numbers with a common base, you add exponents. If I am correct, you need do just a bit more to get the correct answer.