 Exponential inequalities - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: HP Prime (/forum-5.html) +--- Thread: Exponential inequalities (/thread-4903.html) Exponential inequalities - Aries - 10-09-2015 10:54 AM Hi there, it's Aries here ;-) I was checking some exponential function inequalities: solve(3root(2^(4*x-1))>=(2^(x-5))/(4root(16^x))), the calculator reboots. Typing solve(2^((4*x-1)/3)>=(2^(x-5))/(16^(x/4))), all is ok, the calculator returns the correct answer (x>=(-7/2)). Another example: solve(2^(3*x-1)+2^(6*x-2)-2^(3*x+3)<(4^(3*x-2))), 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 ;-) RE: Exponential inequalities - roadrunner - 10-10-2015 10:35 AM 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 RE: Exponential inequalities - parisse - 10-10-2015 11:58 AM The solver can solve polynomial-like 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 pre-simplification command. Here you can do a:=tsimplify(2^(3*x-1)+2^(6*x-2)-2^(3*x+3)-(4^(3*x-2)) This will express the inequation in terms of the minimum possible independant "variables", here 1, hence the inequation becomes polynomial-like, 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! RE: Exponential inequalities - Tim Wessman - 10-10-2015 01:54 PM (10-09-2015 10:54 AM)Aries Wrote:  I was checking some exponential function inequalities: solve(3root(2^(4*x-1))>=(2^(x-5))/(4root(16^x))), the calculator reboots. Is that "root" == nthroot? 3rd root of (2^4*x-1) ? RE: Exponential inequalities - roadrunner - 10-10-2015 11:03 PM Now it all makes sense because: (2^(x-5))/(4 NTHROOT (16^x)) = 1/32 for all x and solve(simplify(3 NTHROOT(2^(4*x-1)))>=1/32) gives the correct answer x>=-7/2 RE: Exponential inequalities - Aries - 10-11-2015 06:39 AM (10-10-2015 01:54 PM)Tim Wessman Wrote:   (10-09-2015 10:54 AM)Aries Wrote:  I was checking some exponential function inequalities: solve(3root(2^(4*x-1))>=(2^(x-5))/(4root(16^x))), the calculator reboots. Is that "root" == nthroot? 3rd root of (2^4*x-1) ? Yes, Tim, it is. This reboots. RE: Exponential inequalities - Aries - 10-11-2015 10:20 AM (10-10-2015 11:58 AM)parisse Wrote:  The solver can solve polynomial-like 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 pre-simplification command. Here you can do a:=tsimplify(2^(3*x-1)+2^(6*x-2)-2^(3*x+3)-(4^(3*x-2)) This will express the inequation in terms of the minimum possible independant "variables", here 1, hence the inequation becomes polynomial-like, 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! Thank you, parisse, calling "tsimplify" does really work fine: However, why that "equal" sign in the solution ? O_o RE: Exponential inequalities - Aries - 10-11-2015 10:23 AM (10-10-2015 11:03 PM)roadrunner Wrote:  Now it all makes sense because: (2^(x-5))/(4 NTHROOT (16^x)) = 1/32 for all x and solve(simplify(3 NTHROOT(2^(4*x-1)))>=1/32) gives the correct answer x>=-7/2 Indeed, nice finding, thank you roadrunner ;-) RE: Exponential inequalities - Aries - 10-17-2015 09:06 AM (10-11-2015 10:20 AM)Aries Wrote:   (10-10-2015 11:58 AM)parisse Wrote:  The solver can solve polynomial-like 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 pre-simplification command. Here you can do a:=tsimplify(2^(3*x-1)+2^(6*x-2)-2^(3*x+3)-(4^(3*x-2)) This will express the inequation in terms of the minimum possible independant "variables", here 1, hence the inequation becomes polynomial-like, 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! Thank you, parisse, calling "tsimplify" does really work fine: However, why that "equal" sign in the solution ? O_o (2^(3*x))/2+(2^(6+x))/4-2^(3*x)*8-(2^(6*x))/16<0; ((2^(6*x)*(4-1))/16)+((2^(3*x)*(1-16))/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*(u-40)<0. u<0 is never verified (2^(3*x)<0). Doing u<40, we've got ln(2^(3*x))