Ambiguous solutions to the same exponential problem - help needed












1












$begingroup$


I came across the below question in an IMO preparatory book and tried to solve it, I followed a certain steps and got a solution, but the book followed a slightly different approach and arrived at a totally different solution, I couldn't find any fault with both approaches.



The problems is as follows:




Find the least value of $a$ for which $5^{1+x}+5^{1-x}$, $a/2$, $25^x+25^{-x}$ are three consecutive terms of an Arithmetic Progression.




My approach :



since the three terms are to be in AP



$a/2 = (5^{1+x}+5^{1-x}+25^x+25^{-x})/2$



$Rightarrow a = 5cdot5^x+5cdot5^{-x}+5^{2x}+5^{-2x}$. Let $5^x = t$ (Step A)



Then $a = 5t+(5/t)+t^2+t^{-2}$. Now let $u=t+1/t$



$Rightarrow a = 5u+u^2-2$, since $u^2 = t^2+t^{-2}+2$ (Step B)



This is a quadratic $u^2+5u-(2+a) = 0$



for $u$ to be real $25+4(2+a) ge 0$,



$Rightarrow a ge -33/4$.



I even tried maximizing the expression on the RHS in Step B , i.e. maximizing $u^2+5u-2$, then differentiating and equating to zero $2u+5 = 0$,



$Rightarrow u = -5/2$, substituting this value in the expression



$(-5/2)^2+5(-5/2)-2 = -33/4$ Same solution as above



Approach followed in the book below.



It's the same up to step A mentioned above.



then the equation becomes



$a = (t-(1/t))^2 +5(sqrt{t}-1/sqrt{t}))^2 +12$



since there are two square expressions, the value of this expression is always $ge12$, hence $age12$.



Both approaches seem fine to me, can someone explain the ambiguity?






inequality







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  • 1




    $begingroup$
    In your approach, $t$ can be any positive real number, thus $u$ has to be in $[2,infty]$. So he condition is not that $Delta geq 0$, but: there is a solution that is $geq 2$.
    $endgroup$
    – Mindlack
    Jan 6 at 9:44










  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11
















1












$begingroup$


I came across the below question in an IMO preparatory book and tried to solve it, I followed a certain steps and got a solution, but the book followed a slightly different approach and arrived at a totally different solution, I couldn't find any fault with both approaches.



The problems is as follows:




Find the least value of $a$ for which $5^{1+x}+5^{1-x}$, $a/2$, $25^x+25^{-x}$ are three consecutive terms of an Arithmetic Progression.




My approach :



since the three terms are to be in AP



$a/2 = (5^{1+x}+5^{1-x}+25^x+25^{-x})/2$



$Rightarrow a = 5cdot5^x+5cdot5^{-x}+5^{2x}+5^{-2x}$. Let $5^x = t$ (Step A)



Then $a = 5t+(5/t)+t^2+t^{-2}$. Now let $u=t+1/t$



$Rightarrow a = 5u+u^2-2$, since $u^2 = t^2+t^{-2}+2$ (Step B)



This is a quadratic $u^2+5u-(2+a) = 0$



for $u$ to be real $25+4(2+a) ge 0$,



$Rightarrow a ge -33/4$.



I even tried maximizing the expression on the RHS in Step B , i.e. maximizing $u^2+5u-2$, then differentiating and equating to zero $2u+5 = 0$,



$Rightarrow u = -5/2$, substituting this value in the expression



$(-5/2)^2+5(-5/2)-2 = -33/4$ Same solution as above



Approach followed in the book below.



It's the same up to step A mentioned above.



then the equation becomes



$a = (t-(1/t))^2 +5(sqrt{t}-1/sqrt{t}))^2 +12$



since there are two square expressions, the value of this expression is always $ge12$, hence $age12$.



Both approaches seem fine to me, can someone explain the ambiguity?






inequality







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    In your approach, $t$ can be any positive real number, thus $u$ has to be in $[2,infty]$. So he condition is not that $Delta geq 0$, but: there is a solution that is $geq 2$.
    $endgroup$
    – Mindlack
    Jan 6 at 9:44










  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11














1












1








1





$begingroup$


I came across the below question in an IMO preparatory book and tried to solve it, I followed a certain steps and got a solution, but the book followed a slightly different approach and arrived at a totally different solution, I couldn't find any fault with both approaches.



The problems is as follows:




Find the least value of $a$ for which $5^{1+x}+5^{1-x}$, $a/2$, $25^x+25^{-x}$ are three consecutive terms of an Arithmetic Progression.




My approach :



since the three terms are to be in AP



$a/2 = (5^{1+x}+5^{1-x}+25^x+25^{-x})/2$



$Rightarrow a = 5cdot5^x+5cdot5^{-x}+5^{2x}+5^{-2x}$. Let $5^x = t$ (Step A)



Then $a = 5t+(5/t)+t^2+t^{-2}$. Now let $u=t+1/t$



$Rightarrow a = 5u+u^2-2$, since $u^2 = t^2+t^{-2}+2$ (Step B)



This is a quadratic $u^2+5u-(2+a) = 0$



for $u$ to be real $25+4(2+a) ge 0$,



$Rightarrow a ge -33/4$.



I even tried maximizing the expression on the RHS in Step B , i.e. maximizing $u^2+5u-2$, then differentiating and equating to zero $2u+5 = 0$,



$Rightarrow u = -5/2$, substituting this value in the expression



$(-5/2)^2+5(-5/2)-2 = -33/4$ Same solution as above



Approach followed in the book below.



It's the same up to step A mentioned above.



then the equation becomes



$a = (t-(1/t))^2 +5(sqrt{t}-1/sqrt{t}))^2 +12$



since there are two square expressions, the value of this expression is always $ge12$, hence $age12$.



Both approaches seem fine to me, can someone explain the ambiguity?






inequality







share|cite|improve this question











$endgroup$




I came across the below question in an IMO preparatory book and tried to solve it, I followed a certain steps and got a solution, but the book followed a slightly different approach and arrived at a totally different solution, I couldn't find any fault with both approaches.



The problems is as follows:




Find the least value of $a$ for which $5^{1+x}+5^{1-x}$, $a/2$, $25^x+25^{-x}$ are three consecutive terms of an Arithmetic Progression.




My approach :



since the three terms are to be in AP



$a/2 = (5^{1+x}+5^{1-x}+25^x+25^{-x})/2$



$Rightarrow a = 5cdot5^x+5cdot5^{-x}+5^{2x}+5^{-2x}$. Let $5^x = t$ (Step A)



Then $a = 5t+(5/t)+t^2+t^{-2}$. Now let $u=t+1/t$



$Rightarrow a = 5u+u^2-2$, since $u^2 = t^2+t^{-2}+2$ (Step B)



This is a quadratic $u^2+5u-(2+a) = 0$



for $u$ to be real $25+4(2+a) ge 0$,



$Rightarrow a ge -33/4$.



I even tried maximizing the expression on the RHS in Step B , i.e. maximizing $u^2+5u-2$, then differentiating and equating to zero $2u+5 = 0$,



$Rightarrow u = -5/2$, substituting this value in the expression



$(-5/2)^2+5(-5/2)-2 = -33/4$ Same solution as above



Approach followed in the book below.



It's the same up to step A mentioned above.



then the equation becomes



$a = (t-(1/t))^2 +5(sqrt{t}-1/sqrt{t}))^2 +12$



since there are two square expressions, the value of this expression is always $ge12$, hence $age12$.



Both approaches seem fine to me, can someone explain the ambiguity?






inequality




inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Jan 6 at 9:53









egreg

185k1486208




185k1486208










asked Jan 6 at 9:34









Madavan ViswanathanMadavan Viswanathan

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  • 1




    $begingroup$
    In your approach, $t$ can be any positive real number, thus $u$ has to be in $[2,infty]$. So he condition is not that $Delta geq 0$, but: there is a solution that is $geq 2$.
    $endgroup$
    – Mindlack
    Jan 6 at 9:44










  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11














  • 1




    $begingroup$
    In your approach, $t$ can be any positive real number, thus $u$ has to be in $[2,infty]$. So he condition is not that $Delta geq 0$, but: there is a solution that is $geq 2$.
    $endgroup$
    – Mindlack
    Jan 6 at 9:44










  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11








1




1




$begingroup$
In your approach, $t$ can be any positive real number, thus $u$ has to be in $[2,infty]$. So he condition is not that $Delta geq 0$, but: there is a solution that is $geq 2$.
$endgroup$
– Mindlack
Jan 6 at 9:44




$begingroup$
In your approach, $t$ can be any positive real number, thus $u$ has to be in $[2,infty]$. So he condition is not that $Delta geq 0$, but: there is a solution that is $geq 2$.
$endgroup$
– Mindlack
Jan 6 at 9:44












$begingroup$
ok ,got it , Thanks very much.
$endgroup$
– Madavan Viswanathan
Jan 6 at 10:11




$begingroup$
ok ,got it , Thanks very much.
$endgroup$
– Madavan Viswanathan
Jan 6 at 10:11










1 Answer
1






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oldest

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1












$begingroup$

The expression $u=t+t^{-1}$ cannot take any value. First of all, $t=5^x>0$; the minimum of $u=t+1/t$ is taken at $t=1$ (AM-GM); hence $t+1/tge2$. All values $ge2$ for $u$ can be obtained.



You then have
$$
a=u^2+5u-2ge 4+10-2=12
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11












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1 Answer
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1 Answer
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active

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active

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active

oldest

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1












$begingroup$

The expression $u=t+t^{-1}$ cannot take any value. First of all, $t=5^x>0$; the minimum of $u=t+1/t$ is taken at $t=1$ (AM-GM); hence $t+1/tge2$. All values $ge2$ for $u$ can be obtained.



You then have
$$
a=u^2+5u-2ge 4+10-2=12
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11
















1












$begingroup$

The expression $u=t+t^{-1}$ cannot take any value. First of all, $t=5^x>0$; the minimum of $u=t+1/t$ is taken at $t=1$ (AM-GM); hence $t+1/tge2$. All values $ge2$ for $u$ can be obtained.



You then have
$$
a=u^2+5u-2ge 4+10-2=12
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11














1












1








1





$begingroup$

The expression $u=t+t^{-1}$ cannot take any value. First of all, $t=5^x>0$; the minimum of $u=t+1/t$ is taken at $t=1$ (AM-GM); hence $t+1/tge2$. All values $ge2$ for $u$ can be obtained.



You then have
$$
a=u^2+5u-2ge 4+10-2=12
$$






share|cite|improve this answer









$endgroup$



The expression $u=t+t^{-1}$ cannot take any value. First of all, $t=5^x>0$; the minimum of $u=t+1/t$ is taken at $t=1$ (AM-GM); hence $t+1/tge2$. All values $ge2$ for $u$ can be obtained.



You then have
$$
a=u^2+5u-2ge 4+10-2=12
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 6 at 9:58









egregegreg

185k1486208




185k1486208












  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11


















  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11
















$begingroup$
ok ,got it , Thanks very much.
$endgroup$
– Madavan Viswanathan
Jan 6 at 10:11




$begingroup$
ok ,got it , Thanks very much.
$endgroup$
– Madavan Viswanathan
Jan 6 at 10:11


















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