20180308, 09:07  #1 
Feb 2018
2^{5}·3 Posts 
How to solve: k(b^m)z=n*d. ?
How to solve: k(b^m)z=n*d. ?
With know values for k,b,z,n. 
20180308, 11:35  #2 
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 

20180308, 15:17  #3 
Aug 2006
3·1,993 Posts 
Are the variables integers, real numbers, or something else?
Are you looking for one solution or a parametrization of all solutions? (Of course there may be no solutions.) 
20180308, 17:46  #4 
Feb 2018
2^{5}×3 Posts 
Type? Integer, of course.
Solutions? The first m. 
20180308, 18:20  #5 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2·19·59 Posts 

20180308, 19:15  #6 
Feb 2018
2^{5}·3 Posts 
and you get something with log(), pi, i, etc.
Not. Solve: 29(5^m)11 = 13*D 
20180308, 19:27  #7 
"Rashid Naimi"
Oct 2015
Remote to Here/There
4302_{8} Posts 

20180308, 19:50  #8 
Feb 2018
2^{5}×3 Posts 
Not.
29(5^M)11=13D. 13 + 11 = (5^0)*24 13 + 24 = (5^0)*37 13 + 37 = (5^2)*2 13 + 2 = (5^1)*3 13 + 3 = (5^0)*16 13 + 16 = (5^0)*29. exponents: 0,0,2,1,0,0. M = sum exponents = 3. D = 1+1*1+1*1*1+1*1*1*25+1*1*1*25*5+1*1*1*25*5*1= 278. Answer: 29(5^3)11=13*278. And 3 is the min value. Is the same thing im posting all the month. 
20180309, 00:46  #9 
"Rashid Naimi"
Oct 2015
Remote to Here/There
4302_{8} Posts 
How do you get
1+1*1+1*1*1+1*1*1*25+1*1*1*25*5+1*1*1*25*5*1 in D = 1+1*1+1*1*1+1*1*1*25+1*1*1*25*5+1*1*1*25*5*1= 278. Thanks in advance. 
20180309, 01:04  #10  
"Rashid Naimi"
Oct 2015
Remote to Here/There
2242_{10} Posts 
Quote:
1+1*1+1*1*1+1*1*1*25+1*1*1*25*5+1*1*1*25*5*1 and why is there no +1*1*1*25*5*1*1 added to the end? Is the added 1 and missing last addend a general rule or subject to variations? Thank you for the reply. Last fiddled with by a1call on 20180309 at 01:04 

20180309, 01:49  #11 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 

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