Fun math question!!

Discus: SAT/ACT Tests and Test Preparation: July 2004 Archive: Fun math question!!
 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.......