M mtsmahia Member Joined Jun 21, 2008 Messages 284 Gender Male HSC 2010 Nov 17, 2008 #1 Sorry for another question on same topic.. Solve x using logarithms 1) 2^x = 5^x-1
lyounamu Reborn Joined Oct 28, 2007 Messages 9,989 Gender Male HSC N/A Nov 17, 2008 #2 mtsmahia said: Sorry for another question on same topic.. Solve x using logarithms 1) 2^x = 5^x-1 Click to expand... 2^x = 5^(x -1) x = ln2(5^(x-1)) x = (x-1)(ln2(5)) x/(x-1) = ln2(5) x/(x-1) = 2.32.. x = 2.32.... x - 2.32.... 2.32... = 1.32... x x = 1.756...
mtsmahia said: Sorry for another question on same topic.. Solve x using logarithms 1) 2^x = 5^x-1 Click to expand... 2^x = 5^(x -1) x = ln2(5^(x-1)) x = (x-1)(ln2(5)) x/(x-1) = ln2(5) x/(x-1) = 2.32.. x = 2.32.... x - 2.32.... 2.32... = 1.32... x x = 1.756...
lyounamu Reborn Joined Oct 28, 2007 Messages 9,989 Gender Male HSC N/A Nov 17, 2008 #3 lou071 said: log both sides and solve for x Click to expand... Duh. Everyone knows that. It is HOW they come about solving it. How helpful.
lou071 said: log both sides and solve for x Click to expand... Duh. Everyone knows that. It is HOW they come about solving it. How helpful.
M mtsmahia Member Joined Jun 21, 2008 Messages 284 Gender Male HSC 2010 Nov 17, 2008 #4 lou071 said: log both sides and solve for x Click to expand... yeah i know---but how i dont know what to do after-- x log(10)2 = (x-1) log(10)5
lou071 said: log both sides and solve for x Click to expand... yeah i know---but how i dont know what to do after-- x log(10)2 = (x-1) log(10)5
lyounamu Reborn Joined Oct 28, 2007 Messages 9,989 Gender Male HSC N/A Nov 17, 2008 #5 lou071 said: if you know the method, they shouldn't have big problem except silly mistakes Click to expand... Do you know what? Most Maths questions are like that. You can at least start with some formula and stuff but it is HOW part where they get stuck. And for some people, even small problem can stumble them over.
lou071 said: if you know the method, they shouldn't have big problem except silly mistakes Click to expand... Do you know what? Most Maths questions are like that. You can at least start with some formula and stuff but it is HOW part where they get stuck. And for some people, even small problem can stumble them over.
M mtsmahia Member Joined Jun 21, 2008 Messages 284 Gender Male HSC 2010 Nov 17, 2008 #6 lyounamu said: 2^x = 5^(x -1) x = ln2(5^(x-1)) x = (x-1)(ln2(5)) x/(x-1) = ln2(5) x/(x-1) = 2.32.. x = 2.32.... x - 2.32.... 2.32... = 1.32... x x = 1.756... Click to expand... whats In--we have learnt it
lyounamu said: 2^x = 5^(x -1) x = ln2(5^(x-1)) x = (x-1)(ln2(5)) x/(x-1) = ln2(5) x/(x-1) = 2.32.. x = 2.32.... x - 2.32.... 2.32... = 1.32... x x = 1.756... Click to expand... whats In--we have learnt it
lyounamu Reborn Joined Oct 28, 2007 Messages 9,989 Gender Male HSC N/A Nov 17, 2008 #7 mtsmahia said: whats In--we have learnt it Click to expand... ln = log base e. I am sure you have learnt this. Because you can change this: log2(5) to loge(5)/loge(2) (or ln5/ln2) by the change of base law.
mtsmahia said: whats In--we have learnt it Click to expand... ln = log base e. I am sure you have learnt this. Because you can change this: log2(5) to loge(5)/loge(2) (or ln5/ln2) by the change of base law.
