solve x, I have really struggled with this.

30,000=0.1(60,000)^x+0.9(10,000)^x

For the life of me i cant remember my a-level maths!

Any help is very much appreciated!

Ben

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- May 17th 2010, 02:02 AMbojker26Logarithm problem
solve x, I have really struggled with this.

30,000=0.1(60,000)^x+0.9(10,000)^x

For the life of me i cant remember my a-level maths!

Any help is very much appreciated!

Ben - May 17th 2010, 02:29 AMFlacidCelery
Start by taking ln of both sides. Then you can use a property of logarithms to bring x down from its perch.

ln(30000)=o.1x*ln(60000)+0.9x*ln(10000)

Factor out x, and bring some terms together.

ln(30000)/(0.1*ln(60000)+0.9*ln(10000))=2x

x=ln(30000)/2*(o.1*ln(60000)+0.9*ln(10000))

This can further be reduced, but I'll leave that up to you.

Best,

-F - May 17th 2010, 03:35 AMbojker26
Thank you! Just still very confused as to what to do next, I had got to that point but couldn't get it any further, sorry i should have explained in the question.!

Ben - May 17th 2010, 05:18 AMearboth
The

**+**-sign idicates a product in the original equation which isn't there.

The only way I see to solve the equation is using an iterative method like Newton:

Construct a function f:

$\displaystyle f(x)=60,000^x+9*10,000^x-300,000$

and calculate the zeros of this function.

$\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$

I used $\displaystyle x_0 = 1$ and after few steps got a result of $\displaystyle x \approx 1.069634491$ - May 18th 2010, 06:46 AMHallsofIvy
This is incorrect. log(ax+ by) is NOT alog(x)+ blog(y) so $\displaystyle log(60000^{.1x}+ 10000^{.9x}\ne .1xln(60000)+ .9xln(10000)$

Earboth's numerical method is probably the best you can do.

Quote:

Factor out x, and bring some terms together.

ln(30000)/(0.1*ln(60000)+0.9*ln(10000))=2x

x=ln(30000)/2*(o.1*ln(60000)+0.9*ln(10000))

This can further be reduced, but I'll leave that up to you.

Best,

-F