Exponential inequalities

10092015, 10:54 AM
Post: #1




Exponential inequalities
Hi there, it's Aries here ;)
I was checking some exponential function inequalities: solve(3root(2^(4*x1))>=(2^(x5))/(4root(16^x))), the calculator reboots. Typing solve(2^((4*x1)/3)>=(2^(x5))/(16^(x/4))), all is ok, the calculator returns the correct answer (x>=(7/2)). Another example: solve(2^(3*x1)+2^(6*x2)2^(3*x+3)<(4^(3*x2))), the calculator gives the advise "Warning, solving parametric inequation requires assumption on parameters otherwise solutions may be missed. The solutions are UNABLE TO ISOLATE x IN..... Why happens this ? Is it due to some sort of complexity or what ? Is there a "workaround" to solve this kind of inequations ? Thanks all for the explanations, Aries ;) 

10102015, 10:35 AM
Post: #2




RE: Exponential inequalities
Hello Aries,
For your first example 16^x is positive for all x so it has no roots. The calculator should just tell you that though, not reboot??!!??!??. For your third example if you turn off exact mode in the cas you get: {(x≥−115.653167402) AND (x≤−60.6140567646),(x≥−60.6140567646) AND (x≤−41.0658420633),(x≥−41.0658420633) AND (x≤−31.0487108669),(x≥−31.0487108669) AND (x≤−24.9577290453),(x≥−24.9577290453) AND (x≤−20.8624598973),(x≥−20.8624598973) AND (x≤−17.9198156801),(x≥−17.9198156801) AND (x≤−15.7029740933),(x≥−15.7029740933) AND (x<1.77397603163)} The last inequality in the list is correct, but the prime should be able to get an exact result. Maybe it's a bug that will be fixed in the next software update. Best regards, Road 

10102015, 11:58 AM
(This post was last modified: 10102015 12:00 PM by parisse.)
Post: #3




RE: Exponential inequalities
The solver can solve polynomiallike equations/inequations. Unfortunately there is no easy way to reduce all equations/inequations to a polynomial. But most of the time, you can help the CAS by calling the right presimplification command. Here you can do
a:=tsimplify(2^(3*x1)+2^(6*x2)2^(3*x+3)(4^(3*x2)) This will express the inequation in terms of the minimum possible independant "variables", here 1, hence the inequation becomes polynomiallike, then solve(a<0) will return the exact answer. The approx answer is partially wrong because the floats are too small for x negative. Remember: the calc computes very fast but it is stupid, humans do not compute fast but humans know what to do, a typical situation where both can cooperate! 

10102015, 01:54 PM
Post: #4




RE: Exponential inequalities
(10092015 10:54 AM)Aries Wrote: I was checking some exponential function inequalities: solve(3root(2^(4*x1))>=(2^(x5))/(4root(16^x))), the calculator reboots. Is that "root" == nthroot? 3rd root of (2^4*x1) ? TW Although I work for HP, the views and opinions I post here are my own. 

10102015, 11:03 PM
Post: #5




RE: Exponential inequalities
Now it all makes sense because:
(2^(x5))/(4 NTHROOT (16^x)) = 1/32 for all x and solve(simplify(3 NTHROOT(2^(4*x1)))>=1/32) gives the correct answer x>=7/2 

10112015, 06:39 AM
Post: #6




RE: Exponential inequalities  
10112015, 10:20 AM
Post: #7




RE: Exponential inequalities
(10102015 11:58 AM)parisse Wrote: The solver can solve polynomiallike equations/inequations. Unfortunately there is no easy way to reduce all equations/inequations to a polynomial. But most of the time, you can help the CAS by calling the right presimplification command. Here you can do Thank you, parisse, calling "tsimplify" does really work fine: However, why that "equal" sign in the solution ? O_o 

10112015, 10:23 AM
Post: #8




RE: Exponential inequalities  
10172015, 09:06 AM
Post: #9




RE: Exponential inequalities
(10112015 10:20 AM)Aries Wrote:(10102015 11:58 AM)parisse Wrote: The solver can solve polynomiallike equations/inequations. Unfortunately there is no easy way to reduce all equations/inequations to a polynomial. But most of the time, you can help the CAS by calling the right presimplification command. Here you can do (2^(3*x))/2+(2^(6+x))/42^(3*x)*8(2^(6*x))/16<0; ((2^(6*x)*(41))/16)+((2^(3*x)*(116))/2)<0; setting (2^(3*x))=u, we've got (3/16)*(u^2)(15/2)*u<0. Doing lcm, we've got (u^2)40*u<0, then u*(u40)<0. u<0 is never verified (2^(3*x)<0). Doing u<40, we've got ln(2^(3*x))<ln(40) and finally x<(ln(5)+ln(8))/ln(8). I'm wondering why that "equal" sign in the Prime result *_* Best, Aries ;) 

10172015, 04:15 PM
Post: #10




RE: Exponential inequalities
Checking is sometimes done with approx value because exact check would be too costly.


10252015, 10:53 AM
Post: #11




RE: Exponential inequalities  
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