Solve for x to 3 decimal places: 3^2x= 4^(1-3x)

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Solve for x to three decimal places: `3^(2x)=4^(1-3x)` :

Take the logarithm of both sides:

`ln(3^(2x))=ln(4^(1-3x))` Now use `lna^x=xlna` :

`=> 2xln3=(1-3x)ln4`

`=>2ln(3)x+3ln(4)x=ln4`

`=>x(2ln3+3ln4)=ln4`

`=> x=(ln4)/(2ln3+3ln4)`

`=>x~~0.218104292`

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The solution is `x~~0.218`

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`3^(2*0.218)=3^(.436)~~1.614450995`

`4^(1-3*.218)=4^(.346)~~1.615521555`

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Hi Embizze, I was just wondering you you be able to help me with this question? thank :)

“The second harmonic adds clearness and brilliance but nothing else,…The octave can introduce no difference of timbre.

The third harmonic again adds a certain amount of brilliance because of its high pitch, but it also introduces a difference of timbre.” (Jeans, J. 1938 *Science & Music*, Cambridge University Press, London)

If **only one harmonic** is combined with the fundamental, investigate Jeans’ claim.

we know that the fundamental tone is y=0.3sin(80*pi*x) (therefore 2nd --> 0.3sin(160*pi*x), 3rd--> 0.3sin(240*pi*x) etc )

I have attached some graphs that I graphed (fund+2nd har, fund+3rd har etc). when you are answering the question could you please make sure you refer to the graph that I have given. hope that makes sense, let me know if you need anymore info,

thanks :)

Hi, should x be to the power of?can you explain it in a simpler form

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The answer to 3 decimal places is x=0.218; no power.

The last two lines show that when you substitute for x into the original equation the left and right hand sides are approximately equal. They will be closer to each other as you keep more digits for x.

Is there a particular step that you are having a problem with?

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