[ARCHIVED THREAD] - Math Experts my Wife needs your help (Page 1 of 2)
Posted: 11/30/2015 9:40:22 PM EDT
|
She cant figure this out
0=12 ln1.5(1.5^x) + 12 ln0.5(0.5^x) She knows the answer is 0.488 but cant figure how to get there. SHes been working on it for a few hours and I am no good in math. everything she has tried completely eliminates the variable from the equation For the Rules 
|
|
Quoted:
Well, look at other examples: 3*3^2 = 3^3 x*x^7 = x^8 Can she see how a*a^x = a^(x+1)? If so, just replace the a with 1.5. Longer explanation available if needed. She now sees where you got the x+1 I think she is confused because the original is 12 ln1.5(1.5^x) She doesn't think (12)(x+1)ln1.5 would work. Would this be the same as (12x+12)ln1.5? |
|
Quoted:
She now sees where you got the x+1 I think she is confused because the original is 12 ln1.5(1.5^x) She doesn't think (12)(x+1)ln1.5 would work. Would this be the same as (12x+12)ln1.5? Quoted:
Quoted:
Well, look at other examples: 3*3^2 = 3^3 x*x^7 = x^8 Can she see how a*a^x = a^(x+1)? If so, just replace the a with 1.5. Longer explanation available if needed. She now sees where you got the x+1 I think she is confused because the original is 12 ln1.5(1.5^x) She doesn't think (12)(x+1)ln1.5 would work. Would this be the same as (12x+12)ln1.5? Step 1-Divide by the 12 No more 12 |
|
Quoted:
She now sees where you got the x+1 I think she is confused because the original is 12 ln1.5(1.5^x) She doesn't think (12)(x+1)ln1.5 would work. Would this be the same as (12x+12)ln1.5? Quoted:
Quoted:
Well, look at other examples: 3*3^2 = 3^3 x*x^7 = x^8 Can she see how a*a^x = a^(x+1)? If so, just replace the a with 1.5. Longer explanation available if needed. She now sees where you got the x+1 I think she is confused because the original is 12 ln1.5(1.5^x) She doesn't think (12)(x+1)ln1.5 would work. Would this be the same as (12x+12)ln1.5? Hah,12 ln1.5(1.5^x) is indeed different from 12ln(1.5*1.5^x), and explains why .488 wasn't working when I was just checking the answer. so... 0 = 12 Ln(1.5)(1.5^x) + 12 Ln(.5)(.5^x) Divide each side by 12, and write xs in front for clarity... 0 = (1.5^x)*Ln(1.5) + (.5^x)*Ln(.5) Subtract (1.5^x)*Ln(1.5) -(1.5^x)*Ln(1.5) = (.5^x)*Ln(.5) Divide to get like terms, Ln and exponent -(1.5^x)/(.5^x) = Ln(.5)/Ln(1.5) (1.5^x)/(.5^x) = -Ln(.5)/Ln(1.5) (1.5/.5)^x = -Ln(.5)/Ln(1.5) 3^x = blah Calculator time from there. .488 actually works in this one. |
|
Quoted:
Raise everything to e Now it's 1 = 1.5*1.5x+.5*.5x simple from there using algebraic laws of exponents. Almost - you need to be careful after raising the entire right side as a power of e (it looks a little different than what you have). But yeah, this method would work, and it's clever too! |
|
Step 1: Make sure the problem you're trying to get GD to solve *is* the problem you're trying to solve.
A lot of people are interpreting lnx(y) to mean ln(xy) -- for no apparent reason. If by lnx(y) you mean "log base x of y" that's wholly different than either lnx(y) or ln(xy) and does, indeed, make the problem look like some sort of identity. |
|
Quoted: Step 1: Make sure the problem you're trying to get GD to solve *is* the problem you're trying to solve. A lot of people are interpreting lnx(y) to mean ln(xy). If by lnx(y) you mean "log base x of y" that's wholly different than either lnx(y) or ln(xy) and does, indeed, make the problem look like some sort of identity. Exactly. A picture of it written down would be easier for us. This problem is either solved by the exponent laws that CoyoteMcD and I have written or by using a characteristic differential equation. |
|
Quoted: Where the fuck is an E in that question/equation/cluster fuck? Quoted: Quoted: I divided 12 out before raising to e. I cannot tell if the whole part after Ln is part of Ln or if it is multiplied by. Where the fuck is an E in that question/equation/cluster fuck? I added it to both sides. Raising e to the power of Ln divides out both and you're left with Ln's value. Raising e to zero equals 1 because any number raised to zero equals 1. |
|
Quoted: I added it to both sides. Raising e to the power of Ln divides out both and you're left with Ln's value. Raising e to zero equals 1 because any number raised to zero equals 1. Quoted: Quoted: Quoted: I divided 12 out before raising to e. I cannot tell if the whole part after Ln is part of Ln or if it is multiplied by. Where the fuck is an E in that question/equation/cluster fuck? I added it to both sides. Raising e to the power of Ln divides out both and you're left with Ln's value. Raising e to zero equals 1 because any number raised to zero equals 1. |
|
Quoted:
I added it to both sides. Raising e to the power of Ln divides out both and you're left with Ln's value. Raising e to zero equals 1 because any number raised to zero equals 1. Quoted:
Quoted:
Quoted:
I divided 12 out before raising to e. I cannot tell if the whole part after Ln is part of Ln or if it is multiplied by. Where the fuck is an E in that question/equation/cluster fuck? I added it to both sides. Raising e to the power of Ln divides out both and you're left with Ln's value. Raising e to zero equals 1 because any number raised to zero equals 1. Wat ^ wat² |
|
Quoted:
Hah,12 ln1.5(1.5^x) is indeed different from 12ln(1.5*1.5^x), and explains why .488 wasn't working when I was just checking the answer. so... 0 = 12 Ln(1.5)(1.5^x) + 12 Ln(.5)(.5^x) Divide each side by 12, and write xs in front for clarity... 0 = (1.5^x)*Ln(1.5) + (.5^x)*Ln(.5) Subtract (1.5^x)*Ln(1.5) -(1.5^x)*Ln(1.5) = (.5^x)*Ln(.5) Divide to get like terms, Ln and exponent -(1.5^x)/(.5^x) = Ln(.5)/Ln(1.5) (1.5^x)/(.5^x) = -Ln(.5)/Ln(1.5) (1.5/.5)^x = -Ln(.5)/Ln(1.5) 3^x = blah Calculator time from there. .488 actually works in this one. Quoted:
Quoted:
Quoted:
Well, look at other examples: 3*3^2 = 3^3 x*x^7 = x^8 Can she see how a*a^x = a^(x+1)? If so, just replace the a with 1.5. Longer explanation available if needed. She now sees where you got the x+1 I think she is confused because the original is 12 ln1.5(1.5^x) She doesn't think (12)(x+1)ln1.5 would work. Would this be the same as (12x+12)ln1.5? Hah,12 ln1.5(1.5^x) is indeed different from 12ln(1.5*1.5^x), and explains why .488 wasn't working when I was just checking the answer. so... 0 = 12 Ln(1.5)(1.5^x) + 12 Ln(.5)(.5^x) Divide each side by 12, and write xs in front for clarity... 0 = (1.5^x)*Ln(1.5) + (.5^x)*Ln(.5) Subtract (1.5^x)*Ln(1.5) -(1.5^x)*Ln(1.5) = (.5^x)*Ln(.5) Divide to get like terms, Ln and exponent -(1.5^x)/(.5^x) = Ln(.5)/Ln(1.5) (1.5^x)/(.5^x) = -Ln(.5)/Ln(1.5) (1.5/.5)^x = -Ln(.5)/Ln(1.5) 3^x = blah Calculator time from there. .488 actually works in this one. My wife wants to give you a big kiss. she was not writing her exponents properly and it was throwing her. She was close |
|
Quoted:
pumbaajk, aren't you glad you asked ? Quoted:
Quoted:
Quoted:
Quoted:
I divided 12 out before raising to e. I cannot tell if the whole part after Ln is part of Ln or if it is multiplied by. Where the fuck is an E in that question/equation/cluster fuck? I added it to both sides. Raising e to the power of Ln divides out both and you're left with Ln's value. Raising e to zero equals 1 because any number raised to zero equals 1. Hey back off, learnings going on here. |
|
It would, but you can't get a value of X that way, because if you are taking the natual log (ln) then you get a simple ln(1.5^x+1) = -ln (0.5^x+1) which is not solvable for any X. The natural log takes precedence over multiplication, so ln 1.5 * 1.5^x is ln (1.5) * 1.5^x rather than ln (1.5 * 1.5^x)
if instead you have 12 * ln (1.5) * 1.5^x + 12 * ln 0.5 * (0.5^x) = 0 then you can solve: 12 * ln (1.5) * 1.5^x = -12 * ln (0.5) * (0.5^x) Ln (1.5) * (1.5^x) = -ln (0.5) * (0.5^x) ** get rid of the constant 12 on both sides 1.5^x / 0.5^x = - ln (0.5)/ln(1.5) ** move the X and ln terms to opposite sides by dividing both sides by ln (1.5) and (0.5)^x 3^x = -ln (0.5) / ln (1.5) ** simplify the X term - (1.5/.5)^x 3^x = ln (2) /ln (1.5) ** Ln (1/a) = - ln (a) now evaluate and you get 3^x = ~1.709 now, the natural log of 3^x is x ln 3, so we take the natural log of both sides and get: x ln 3 = ln (1.709) x = ln (1.709) / ln (3) x = .488 It has been a long while since college and I don't see where this equation would represent any process, so presumably it is an abstract problem to examine natural logs. Mike eta - spent way too much time working that, as it has been solved already... |
|
Quoted:
It would, but you can't get a value of X that way, because if you are taking the natual log (ln) then you get a simple ln(1.5^x+1) = -ln (0.5^x+1) which is not solvable for any X. The natural log takes precedence over multiplication, so ln 1.5 * 1.5^x is ln (1.5) * 1.5^x rather than ln (1.5 * 1.5^x) if instead you have 12 * ln (1.5) * 1.5^x + 12 * ln 0.5 * (0.5^x) = 0 then you can solve: 12 * ln (1.5) * 1.5^x = -12 * ln (0.5) * (0.5^x) Ln (1.5) * (1.5^x) = -ln (0.5) * (0.5^x) ** get rid of the constant 12 on both sides 1.5^x / 0.5^x = - ln (0.5)/ln(1.5) ** move the X and ln terms to opposite sides by dividing both sides by ln (1.5) and (0.5)^x 3^x = -ln (0.5) / ln (1.5) ** simplify the X term - (1.5/.5)^x 3^x = ln (2) /ln (1.5) ** Ln (1/a) = - ln (a) now evaluate and you get 3^x = ~1.709 now, the natural log of 3^x is x ln 3, so we take the natural log of both sides and get: x ln 3 = ln (1.709) x = ln (1.709) / ln (3) x = .488 It has been a long while since college and I don't see where this equation would represent any process, so presumably it is an abstract problem to examine natural logs. Mike eta - spent way too much time working that, as it has been solved already... It is, it's for her calc class and they are doing different methods to fine mins and maxs |