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Logarithmic equation with parameter
- Thread starter shervlad
- Start date
D
Deleted member 4993
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I need to find all values of the real parameter "a" for which the equation \(\displaystyle \left| {a - \log x} \right| + \left| {a + \log x} \right| = 2\) has two solutions. Find the solutions.
You need find these solutions - what effort did you make?
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Is such a solution correct?
\(\displaystyle \eqalign{ & \left| {a - \log x} \right| + \left| {a + \log x} \right| = 2 \cr
& \log x = t \cr
& \left| {a - t} \right| + \left| {a + t} \right| = 2 \cr
& \left( {{a^2} - 2at + {t^2}} \right) + 2\left| {\left( {a - t} \right)\left( {a + t} \right)} \right| + \left( {{a^2} + 2at + {t^2}} \right) = 4 \cr
& 2{a^2} + 2{t^2} + 2\left| {{a^2} - {t^2}} \right| = 4 \Leftrightarrow {a^2} + {t^2} + \left| {{a^2} - {t^2}} \right| = 2 \cr
& \left[ \matrix{
{a^2} \ge {t^2},\,\,\,\,\,{a^2} + {t^2} + {a^2} - {t^2} = 2 \cr
{a^2} < {t^2},\,\,\,\,\,{a^2} + {t^2} - {a^2} + {t^2} = 2 l \cr} \right. \Leftrightarrow \left[ \matrix{
2{a^2} = 2 \cr
2{t^2} = 2 \cr} \right. \Leftrightarrow \left[ \matrix{
a = \pm 1 \cr
t = \pm 1 \cr} \right. \cr
& t{\rm{\,\ has\,\ two\,\ solutions \,\ for\,\ }}{a^2} < {t^2} \Leftrightarrow - t < a < t \Leftrightarrow - 1 < a < 1 \Leftrightarrow \cr
& for{\rm{\,\ }}a \in \left( { - 1;1} \right),\,\,t = \pm 1 \Leftrightarrow \log x = \pm 1 \Leftrightarrow {x_1} = 10,\,\,{x_2} = {1 \over {10}} \cr} \)
\(\displaystyle \eqalign{ & \left| {a - \log x} \right| + \left| {a + \log x} \right| = 2 \cr
& \log x = t \cr
& \left| {a - t} \right| + \left| {a + t} \right| = 2 \cr
& \left( {{a^2} - 2at + {t^2}} \right) + 2\left| {\left( {a - t} \right)\left( {a + t} \right)} \right| + \left( {{a^2} + 2at + {t^2}} \right) = 4 \cr
& 2{a^2} + 2{t^2} + 2\left| {{a^2} - {t^2}} \right| = 4 \Leftrightarrow {a^2} + {t^2} + \left| {{a^2} - {t^2}} \right| = 2 \cr
& \left[ \matrix{
{a^2} \ge {t^2},\,\,\,\,\,{a^2} + {t^2} + {a^2} - {t^2} = 2 \cr
{a^2} < {t^2},\,\,\,\,\,{a^2} + {t^2} - {a^2} + {t^2} = 2 l \cr} \right. \Leftrightarrow \left[ \matrix{
2{a^2} = 2 \cr
2{t^2} = 2 \cr} \right. \Leftrightarrow \left[ \matrix{
a = \pm 1 \cr
t = \pm 1 \cr} \right. \cr
& t{\rm{\,\ has\,\ two\,\ solutions \,\ for\,\ }}{a^2} < {t^2} \Leftrightarrow - t < a < t \Leftrightarrow - 1 < a < 1 \Leftrightarrow \cr
& for{\rm{\,\ }}a \in \left( { - 1;1} \right),\,\,t = \pm 1 \Leftrightarrow \log x = \pm 1 \Leftrightarrow {x_1} = 10,\,\,{x_2} = {1 \over {10}} \cr} \)
HallsofIvy
Elite Member
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- Jan 27, 2012
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- 7,763
That strikes me as overly complicated. (Getting rid of absolute values by squaring always does!)
Yes, letting t= ln(x), so that the equation becomes |a- t|+ |a+ t|= 2 is a good idea. Now separate into cases:
1) if a> t> 0, both a- t and a+ t are positive so that becomes a- t+a+t=2a= 2. If a= 1, that has every value of t between 0 and 1 as a solution. Otherwise there is no solution.
2) If t> a> 0, a- t is negative while a+t is positive. The equation becomes -(a- t)+ (a+ t)= 2t= 2 which has the single solution t= 1 for all a between 0 and 1 for all a between 0 and 1.
3) If 0> a> t a+ t is negative while a- t is positive so the equation becomes (a-t)- (a+ t)= -2t= 2 which has the single solution t= -1 for all a between 0 and -1.
4) if 0> t> a, t+ a and a- t are both negative so the equation becomes -(a- t)- (a+ t)= -2a= 2. If a= -1, that has every value of t between 0 and -1 as a solution. Otherwise it has no solution.
Yes, letting t= ln(x), so that the equation becomes |a- t|+ |a+ t|= 2 is a good idea. Now separate into cases:
1) if a> t> 0, both a- t and a+ t are positive so that becomes a- t+a+t=2a= 2. If a= 1, that has every value of t between 0 and 1 as a solution. Otherwise there is no solution.
2) If t> a> 0, a- t is negative while a+t is positive. The equation becomes -(a- t)+ (a+ t)= 2t= 2 which has the single solution t= 1 for all a between 0 and 1 for all a between 0 and 1.
3) If 0> a> t a+ t is negative while a- t is positive so the equation becomes (a-t)- (a+ t)= -2t= 2 which has the single solution t= -1 for all a between 0 and -1.
4) if 0> t> a, t+ a and a- t are both negative so the equation becomes -(a- t)- (a+ t)= -2a= 2. If a= -1, that has every value of t between 0 and -1 as a solution. Otherwise it has no solution.
yeah, I tried that too, but the question is for which volues of the real parameter a, the equation has exactly two solution , and for me it is easier to square the equation (especially if the equation is not very long), and to work with only one absolute value, especially when I have two variables...That strikes me as overly complicated. (Getting rid of absolute values by squaring always does!)
Yes, letting t= ln(x), so that the equation becomes |a- t|+ |a+ t|= 2 is a good idea. Now separate into cases:
1) if a> t> 0, both a- t and a+ t are positive so that becomes a- t+a+t=2a= 2. If a= 1, that has every value of t between 0 and 1 as a solution. Otherwise there is no solution.
2) If t> a> 0, a- t is negative while a+t is positive. The equation becomes -(a- t)+ (a+ t)= 2t= 2 which has the single solution t= 1 for all a between 0 and 1 for all a between 0 and 1.
3) If 0> a> t a+ t is negative while a- t is positive so the equation becomes (a-t)- (a+ t)= -2t= 2 which has the single solution t= -1 for all a between 0 and -1.
4) if 0> t> a, t+ a and a- t are both negative so the equation becomes -(a- t)- (a+ t)= -2a= 2. If a= -1, that has every value of t between 0 and -1 as a solution. Otherwise it has no solution.
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
Well, that question was answered in my "2" and"3".
If a= 0, the equation becomes 2|t|= 2 so there are two solutions, t= 1 and t= -1.
Of course, that is for t. Don't forget that t= ln(x). for a in (-1, 0), ln(x)= -1 so that x= 1/e. For a in (0, 1), ln(x)= 1 so x= e.
As I said above, a= 0 gives two solutions. You might want to check a= -1 and a= 1 separately.
There exist a single solution for all a between -1 and 0 and for a between 0 and 1, not inclusive. If -1< a< 0, t= -1 is the only solution. If 0< a< 1, t= 1 is the only solution.2) If t> a> 0, a- t is negative while a+t is positive. The equation becomes -(a- t)+ (a+ t)= 2t= 2 which has the single solution t= 1 for all a between 0 and 1 for all a between 0 and 1.
3) If 0> a> t a+ t is negative while a- t is positive so the equation becomes (a-t)- (a+ t)= -2t= 2 which has the single solution t= -1 for all a between 0 and -1.
If a= 0, the equation becomes 2|t|= 2 so there are two solutions, t= 1 and t= -1.
Of course, that is for t. Don't forget that t= ln(x). for a in (-1, 0), ln(x)= -1 so that x= 1/e. For a in (0, 1), ln(x)= 1 so x= e.
As I said above, a= 0 gives two solutions. You might want to check a= -1 and a= 1 separately.
Of course, that is for t. Don't forget that t= ln(x).
for a in (-1, 0),
ln(x) = -1 so that x= 1/e. <----
For a in (0, 1), ln(x) = 1 so x= e. <----
Do not assume here that log(x) equals ln(x).
shervlad worked with it as \(\displaystyle \log_{10}(x).\)