How to solve this equation with pari gp ...

[rockzur]

I don't know how to put the solve command on the pari gp command line to solve logarithmic equations. Can somebody help me.

[a1call]

As I was taught by Paul Underwood,
To get log (x) in base n use
log(x)/log(n)
Pari-GP gives the natural log of the number otherwise known as ln(x) by the rest of the world.

[rockzur]

I mean how to solve for example 2 ^ x -5 = -6. Using the solve command

[axn]

? ?solve
solve(X=a,b,expr): real root of expression expr (X between a and b), where expr(a)*expr(b)<=0.

Your sample equation boils down to 2^x+1=0, which doesn't have real roots. Let's try with an equation that does.

2^x-5==0. We need to know an interval (a,b) where a root lies. We know that 2^2 < 5 < 2^3. So let's try (2,3)

? solve(x=2, 3, 2^x-5)
%2 = 2.3219280948873623478703194294893901759

[rockzur]

It would be solve(2^x=8); For example

[axn]

Originally Posted by rockzur

It would be solve(2^x=8); For example

No. You give a real valued expression in one variable, and it will find one real-valued solution where the expression becomes zero. So you recast your equation as 2^x-8=0 and just give the left side of the equality as input.

? solve(t=0,10,2^t-8)
%1 = 3.0000000000000000000000000000000000000

[R. Gerbicz]

Originally Posted by axn

? ?solve
solve(X=a,b,expr): real root of expression expr (X between a and b), where expr(a)*expr(b)<=0.

What is missing here is that you need a continuous function. Otherwise the result is crap:

? solve(x=0,3,if(x<2,-1,1))
%3 = 1.9999999999999999999999999999999999999
?

[rockzur]

What I want to know is how is used solve command.

[paulunderwood]

?solve
solve(X=a,b,expr): real root of expression expr (X between a and b), where 
expr(a)*expr(b)<=0.

For logarithmic equations use pen and paper and an internet search engine.

[R. Gerbicz]

Originally Posted by rockzur

What I want to know is how is used solve command.

Good question, it is not binary search, it should be another root finding algorithm. Btw in some really trivial cases the solve breaks:

? solve(x=-1,2,x^3)
  ***   at top-level: solve(x=-1,2,x^3)
  ***                              ^----
  ***   sorry, solve recovery [too many iterations] is not yet implemented.
  ***   Break loop: type 'break' to go back to GP prompt
break>

What is interesting is that their approx roots was so close to the x=0 solution, don't know why they haven't aborted the search.

? solve(x=-1,2,print(x);x^3)
-1.0000000000000000000000000000000000000
2.0000000000000000000000000000000000000
-0.66666666666666666666666666666666666667
-0.53132832080200501253132832080200501253
-0.39585716304753417148752640416134559761
0.80207141847623291425623679791932720119
-0.26729770261219306942169777701737070320
...
-4.1842288737629149621829297627585343823 E-26
8.5184705351226720431227968573430956740 E-26
-2.8383161799160252299934361727339277773 E-26
  ***   at top-level: solve(x=-1,2,print(x);x^3)
  ***                              ^-------------
  ***   sorry, solve recovery [too many iterations] is not yet implemented.
  ***   Break loop: type 'break' to go back to GP prompt

[jane477]

Good question, it is not binary search, it should be another root finding algorithm. Btw in some really trivial cases the solve breaks:

? solve(x=-1,2,x^3)
  ***   at top-level: solve(x=-1,2,x^3)
  ***                              ^----
  ***   sorry, solve recovery [too many iterations] is not yet implemented.
  ***   Break loop: type 'break' to go back to GP prompt
break>

What is interesting is that their approx roots was so close to the x=0 solution, don't know why they haven't aborted the search.

? solve(x=-1,2,print(x);x^3)
-1.0000000000000000000000000000000000000
2.0000000000000000000000000000000000000
-0.66666666666666666666666666666666666667
-0.53132832080200501253132832080200501253
-0.39585716304753417148752640416134559761
0.80207141847623291425623679791932720119
-0.26729770261219306942169777701737070320
...
-4.1842288737629149621829297627585343823 E-26
8.5184705351226720431227968573430956740 E-26
-2.8383161799160252299934361727339277773 E-26
  ***   at top-level: solve(x=-1,2,print(x);x^3)
  ***                              ^-------------
  ***   sorry, solve recovery [too many iterations] is not yet implemented.
  ***   Break loop: type 'break' to go back to GP prompt

what an elegant solution