B bojker26 May 2010 2 0 May 17, 2010 #1 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

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

F FlacidCelery May 2010 7 3 May 17, 2010 #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 Reactions: bojker26

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

B bojker26 May 2010 2 0 May 17, 2010 #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

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

earboth MHF Hall of Honor Jan 2006 5,854 2,553 Germany May 17, 2010 #4 FlacidCelery said: 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) Click to expand... The +-sign idicates a product in the original equation which isn't there. bojker26 said: 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 Click to expand... 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 \(\displaystyle x \approx 1.069634491\) Reactions: HallsofIvy

FlacidCelery said: 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) Click to expand... The +-sign idicates a product in the original equation which isn't there. bojker26 said: 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 Click to expand... 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 \(\displaystyle x \approx 1.069634491\)

H HallsofIvy MHF Helper Apr 2005 20,249 7,909 May 18, 2010 #5 FlacidCelery said: 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) Click to expand... 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 Click to expand...

FlacidCelery said: 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) Click to expand... 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 Click to expand...