# Math Help - Logarithm problem

1. ## Logarithm 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

2. 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

3. 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

4. Originally Posted by FlacidCelery
Start by taking ln of both sides. <<< better not!
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)
The +-sign idicates a product in the original equation which isn't there.

Originally Posted by bojker26
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
The only way I see to solve the equation is using an iterative method like Newton:

Construct a function f:

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

and calculate the zeros of this function.

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

I used $x_0 = 1$ and after few steps got a result of $x \approx 1.069634491$

5. Originally Posted by FlacidCelery
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)
This is incorrect. log(ax+ by) is NOT alog(x)+ blog(y) so $log(60000^{.1x}+ 10000^{.9x}\ne .1xln(60000)+ .9xln(10000)$

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