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
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
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 $\displaystyle x \approx 1.069634491$
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.
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