Melody

1220*5311=6479420^how to get ans 4947

1220*5311 = 6479420

6479420^x=4947

ln(6479420^x)=ln(4947)

xln(6479420)=ln(4947)

x=ln(4947)/ln(6479420)

x= ln(4947)/ln(6479420) = 0.5423654574711226

ok

\(x=u^2+1\\ dx=2u\;\;du\\~\\ x-1=u^2\qquad u^2\ge0\;\;so\;\; x\ge1 \\ u=\pm\sqrt{x-1}\;\; \\ \text {Rather than having }x=u^2+1\\\text{I am going to have }u=+\sqrt{x-1 }\\ \text{Everything else is the same but I don't have to worry about the negative posibility}\\ \)

\(\displaystyle \int \frac{1}{2x\sqrt{x+1}}\;dx\\ =\displaystyle \int \frac{1}{2(u^2-1)u}\;2u\;\;du\\ =\displaystyle \int \frac{1}{u^2-1}\;\;du\\ =-\displaystyle \int \frac{1}{1-u^2}\;\;du\\ =-tanh^{-1}(u)+c\\ \qquad\text{equivalent for restricted values (from Wofram|alpha) }\\ \)

\(=\frac{1}{2} \left[ ln\frac{1-u}{1+u} \right]+c\\ =\frac{1}{2} \left[ ln\frac{1-u}{1+u} \right]+c\\ =\frac{1}{2} \left[ ln\frac{u-1}{u+1} \right]+c\\ \qquad \text{ Now c can be replaced with } logA\\ \qquad \text{Where A is a constant >0}\\ \qquad \text{I do not know why this condition was not given...}\\ =ln \left[ \frac{u-1}{u+1} \right]^{1/2}+lnA\\ =ln\left[A \left[ \frac{u-1}{u+1} \right]^{1/2}\right]\\ =ln\left[A\sqrt{ \left[ \frac{u-1}{u+1} \right]}\;\right]\\ \qquad u=\sqrt{x+1}\\ =ln\left[\;A\sqrt{ \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} }\;\;\;\right]\\ \)

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No it is not!

I do not think it is what you do!

I think you try your best to learn from all the answers you are given!

I doubt very much that felrose has any idea what you have done Chris.

I doubt if he/she is even interested.

He/she is not using us to learn he is just using us to do his/her homework.

Hi 8MP

I have played with this question a lot.  

I have learned to use GeoGebra better which I am quite pleased about but I am no closer to a solution.

I have aske more senior mathematicians to help. So far we have nothing.

I did get a response from someone that i have great respect for.  

He said he toyed with it in Matematica and come up with zilch.  

Is it possible that there is an error in the question?  Maybe it is not doable...

What have you tried so far?    

Hi Felrode13

You have been submitting a lot of questions.

I would like to start seeing what you can do for yourself.  You should not just be copying our work.

You will not learn any thing that way.

Could you please show me what you have attempted yourself to solve this problem?

Also, are you meant to be working in degrees or in radians?

1+2log(x,3)=log(28,3) + log(x,3)- log(9,3)

\(1+2log_3x=log_328 + log_3x- log_39\\ 1+2log_3x=log_328 + log_3x- 2\\ log_3 x=-3+log_3 28\\ x=3^{-3+log_3 28 }\\\)

3^(-3+(log(28)/log(3))) = 1.037037037037037

x=1.037037037037037

Hi Alan :))