Any help would be appreciated. :)
I tried splitting the equation about $x=3$, but the terms $x^2$ and $2^x$ Together in the equation(s) are troubling me.
I don't know why I'm unable to apply the property $log_ax=\frac1{log_xa}$
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Harshal Gajjar
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+1 Thanks SDiv, but is it correct to say that there exist no roots<3? (Which we would get, if there are, after solving for $x<3$? – Harshal Gajjar Oct 17 '14 at 10:45
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Beside $x=3$, there are two roots to the equation. They are close to $\pm \frac{1}{2}$. Just find them following Georg's answer. Cheers :-) – Claude Leibovici Oct 17 '14 at 10:59
1 Answers
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HINT
I would say
$x^2 2^{x+1} + 2^{|x-3|+2} = x^2 2^{|x-3|+4} + 2^{x-1}$ |*2
$4x^2 2^{x} + 2^{|x-3|+3} = 4x^2 2^{|x-3|+3} + 2^{x}$
$4x^2 (2^{x} -2^{|x-3|+3})-(2^{x} -2^{|x-3|+3}) = 0$
$(4x^2 -1) (2^{x} -2^{|x-3|+3}) = 0$
georg
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Thanks for the help, Georg. But how to solve further when $2^x$ and $x^2$ arrive in the equation together? – Harshal Gajjar Oct 17 '14 at 10:44
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Note that you have a product equal to zero so that one or the other or both terms must be zero. – SDiv Oct 17 '14 at 10:47
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Another way to do it would have been casewise, case 1 assume $x\geq3$ and case 2 assume $x\leq3$ in order to remove the absolute value sign. – SDiv Oct 17 '14 at 10:49
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If you replace $x$ by $y^2+3$, you arrive to an identity for any value of $y$. Then, just work the case $x \lt 3$ since $x=3$ is a solution. – Claude Leibovici Oct 17 '14 at 10:52
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For $x<3$ you have the solution $x=\pm\frac{1}{2}$ from $4x^2-1=0$. – gammatester Oct 17 '14 at 11:01