Logarithm question

[sayo]

 

[Deleted member 4993]

View attachment 26853

assume
x^[ln(y)] = m .... then

ln(m) = ??

Please share your thoughts/work about your assignment.

 

[Dr.Peterson]

Interesting problem. Where does it come from?

I suspect you may have omitted some information, because I find that ln(x)ln(y) is not constant in general. It will be constant under some conditions.

Please show us the entire problem as given to you, along with whatever work you have done.

 

[lex]

Write [MATH]x[/MATH] as [MATH]e^{lnx}[/MATH] and [MATH]y[/MATH] as [MATH]e^{lny}[/MATH] and you will be able to get an expression for [MATH]e^{lnx \cdot lny}[/MATH] and therefore from that an expression for [MATH]lnx \cdot lny[/MATH]
(You could leave the [MATH]x^{lnx}[/MATH] as it is. There is no benefit in changing it).

 

[sayo]

assume
x^[ln(y)] = m .... then

ln(m) = ??

Please share your thoughts/work about your assignment.

ln(y).ln(x) I tried to get the ln of both sides but I'm stuck

Interesting problem. Where does it come from?

I suspect you may have omitted some information, because I find that ln(x)ln(y) is not constant in general. It will be constant under some conditions.

Please show us the entire problem as given to you, along with whatever work you have done.

This question was asked in the high school test book in Turkey.

Write [MATH]x[/MATH] as [MATH]e^{lnx}[/MATH] and [MATH]y[/MATH] as [MATH]e^{lny}[/MATH] and you will be able to get an expression for [MATH]e^{lnx \cdot lny}[/MATH] and therefore from that an expression for [MATH]lnx \cdot lny[/MATH]
(You could leave the [MATH]x^{lnx}[/MATH] as it is. There is no benefit in changing it).

I coudn't finish it. Can you help me please?

 

[Dr.Peterson]

View attachment 26858
This question was asked in the high school test book in Turkey.


I coudn't finish it. Can you help me please?

Thanks, but please translate the words! I can't tell yet whether they affect the answer.

Also, when a problem includes choices, they need to be included; sometimes the list is an essential part of the question. In this case, it shows at least that they expect a specific number, not an expression as lex suggested.

 

[sayo]

Thanks, but please translate the words! I can't tell yet whether they affect the answer.

Also, when a problem includes choices, they need to be included; sometimes the list is an essential part of the question. In this case, it shows at least that they expect a specific number, not an expression as lex suggested.

Since **equation** , where x and y are positive real numbers, which of the following is the equivalent of lnx.lny?

 

[lex]

In this case, it shows at least that they expect a specific number, not an expression as lex suggested.

It is not Lex who suggests it; the question implies it.

The question is wrong. It should read:
[MATH]x^{lny}+y^{lnx}+x^{\boldsymbol{lny}} \cdot y^{lnx} =35[/MATH]and can be solved as I suggested.
(The answer is (B) ln5)

 

[Steven G]

I too get that lnx*lny is not a constant.

 

[Steven G]

Yes, I too get lnx*lny = ln5. It follows from the definition of logs.

 

[lex]

I coudn't finish it. Can you help me please?

You may try the corrected question:

(The question is wrong). It should read:
[MATH]x^{lny}+y^{lnx}+x^{\boldsymbol{lny}} \cdot y^{lnx} =35[/MATH]
(The answer is (B) ln5)

Write [MATH]x[/MATH] as [MATH]e^{lnx}[/MATH] and [MATH]y[/MATH] as [MATH]e^{lny}[/MATH] and you will be able to get an expression for [MATH]e^{lnx \cdot lny}[/MATH] and therefore from that an expression for [MATH]lnx \cdot lny[/MATH]

If there still is a problem completing it, if you would like to post your work we can take a look at it and get you to the answer.

 

[sayo]

Thank you all.