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By Thermodude (Thermodude) on Tuesday, July 13, 2004 - 09:36 pm: Edit |

Just want to see how quick CC'ers are with this fun math question:

Find the solutions to the following equation:

3^x = x + 1 + [log (2x + 1)]/log 3

By Energy1 (Energy1) on Tuesday, July 13, 2004 - 09:54 pm: Edit |

x = 1

By Etsrep78328 (Etsrep78328) on Tuesday, July 13, 2004 - 10:32 pm: Edit |

The correct solutions are x = 0 and x = 1.

By Thermodude (Thermodude) on Tuesday, July 13, 2004 - 11:17 pm: Edit |

Just curious...how did you guys go about solving the problem? Just curious to see if anyone solved it using the algebraic method.

By Energy1 (Energy1) on Tuesday, July 13, 2004 - 11:51 pm: Edit |

Oops. I suspected there would be a second solution but I was too lazy to check

I multiplied both sides by log 3. Since, both sides will then be in terms of log, you can change it into a much simpler equation using rules of log.

There are many ways to do a problem so always look for a simpler solution.

By Al0 (Al0) on Wednesday, July 14, 2004 - 12:14 am: Edit |

I'm pretty sure it is impossible to solve this problem algebraicly (id est isolate x). Pursuing Energy1's line:

3^x=x+1+log(2x+1)/log(3)

(3^x)*log(3)=(x+1)*log(3)+log(2x+1)

log(3^[3^x])=log(3^[x+1])+log(2x+1)

log(3^[3^x])=log(3^[x+1]*[2x+1])

3^(3^x)=(2x+1)(3^[x+1])

3^(3^x-x-1)=2x+1

This obviously can't be solved because it is impossible to get all the x's either in or out of an exponent/logarithm. I await your response! What is the algebraic method you speak of?

By Devious (Devious) on Wednesday, July 14, 2004 - 01:45 pm: Edit |

Completing Al0's solution:

3^(3^x-x-1)=2x+1=(2x+1)^1

3=2x+1 & 3^x-x-1=1

This gives x=1.

The quickest method to obtain this answer is to realise that log(2x+1)/log3 would equal 1 if x=1, and then substituting this value of x in the entire equation to see if it works out.

And to find the other solution, you'd need to connect log(2x+1) to log1(=0), i.e. x=0.

By Thermodude (Thermodude) on Wednesday, July 14, 2004 - 05:34 pm: Edit |

yeah...i used a different method....I just let t = log (2x + 1) / log 3....then i had

3^x = x + 1 + t...and

3^t = 2x + 1....

..pair of simeltaneous equations....combine 'em together to get...

3^x - 3^t = -x + t

3^x + x = 3^t + t

...therefore...x = t

..so 3^x = 2x + 1...and

..1 = 2x/3^x + 1/3^x...and from there you can see that x = 0 and 1...

..anyways..

By Al0 (Al0) on Wednesday, July 14, 2004 - 06:05 pm: Edit |

Devious, given a^b=c^d, you can't just assume that a=c and b=d. For instance, if you have

3^(x/4)=(2x-6)^log(9) you would suggest that x/4=log(9) and (2x-6)=3 which yield x=4log(9) and x=9/2, which yield 2.853=1.598 and 3.442=2.853 -- neither of which is true. The actual solutions are 5.296 and 8. I still think that the original equation cannot be solved with algebra, but is easily eye-balled.

Thermodude -

You still never actually solved the problem, just got it into another form which I don't see as much better for solving intuitively than the original.

By Tongos (Tongos) on Saturday, July 17, 2004 - 12:08 pm: Edit |

look at what i have started.......

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