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#1 2006-11-26 04:08:34

Thatcher
Guest

Logs

Hey guys

How would you go about solving 2^x+1=5^x

#2 2006-11-26 18:56:39

Toast
Real Member
Registered: 2006-10-08
Posts: 1,321

Re: Logs

Hi Thatcher, I just asked one of my friends and he said it is unsolvable using logs. He said you need to use calculus, which I haven't learnt.
Either by graphing or using a calculator I got x ≈ 0.563895524260.
A forum member should come along soon and help you with the method though, don't worry.

Last edited by Toast (2006-11-26 18:58:37)

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#3 2006-11-26 19:24:34

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: Logs

yes, i tried with logs, but it is unsolvable via them, im not sure how you would apply calculus though, maybe youre friend could give a bit more help tongue


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#4 2006-11-26 19:32:50

Toast
Real Member
Registered: 2006-10-08
Posts: 1,321

Re: Logs

Well, I can't ask him again till Thursday hmm.

Try just analysing the graphs of


and

see if you can come up with something

Last edited by Toast (2006-11-26 19:50:19)

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#5 2006-11-27 01:43:38

Thatcher
Guest

Re: Logs

I solved it using logs.

2^(x+1)=5^x

= 2^x * 2 = 5^x

= 2 = (5/2)^x

= log2 = x log (5/2)

therefore x = 0.756

#6 2006-11-27 02:02:54

Toast
Real Member
Registered: 2006-10-08
Posts: 1,321

Re: Logs

Oh dear, you didn't include the brackets in the first question? tongue
Oh well, good job on completing the question.;)

P.S. I'll still try and get back to you on

Last edited by Toast (2006-11-27 02:10:00)

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#7 2006-11-27 03:48:12

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: Logs

those brackets make all the difference tongue


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