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0
563
2
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$$\left({\left({\frac{\left({\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}}\right)}^{{log8}{\left({\mathtt{0.25}}\right)}}\right) = {{\mathtt{1}}}^{{{log}}_{{\mathtt{8}}}{\left({\mathtt{0.25}}\right)}}$$

$${{\mathtt{2}}}^{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{\mathtt{\,-\,}}{{\mathtt{4}}}^{{\mathtt{x}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{8}}}^{{\mathtt{x}}} = {\mathtt{0}}$$

 

  Can anyone tell me what are the steps for this ?

  And how i can solve the second one ?

 Nov 19, 2014

Best Answer 

 #2
avatar+95177 
+10

$$\\2^{(x-2)}-4^x+8^x=0\\
2^x*2^{-2}-2^{2x}+2^{3x}=0\\
2^x*2^{-2}-(2^{x})^2+(2^{x})^3=0\\
2^x(2^{-2}-2^{x}+(2^{x})^2)=0\\
2^x \mbox{ cannot equal zero so}\\
2^{-2}-2^{x}+(2^{x})^2=0\\
$Let $ y=2^x\\
2^{-2}-y+y^2=0\\
y^2-y+\frac{1}{4}=0\\\\
y=\frac{1\pm\sqrt{1-1}}{2}=\frac{1}{2}\\
$therefore$\\
\frac{1}{2}=2^x\\
2^{-1}=2^x\\
x=-1$$

.
 Nov 20, 2014
 #1
avatar+17747 
+5

For the top equation:  there is: (√2 - 1) / (√2 - 1)   --->  but this equals 1. So any finite exponent of 1 will work.

For the second equation:

2^(x - 2) - 4^x + 8^x  =  0

--->  2^(x - 2) - (2^2)^x + (2^3)^x  =  0

--->  2^(x - 2) - 2^(2x) + 2^(3x)  =  0     --->  x = -1     (I solved it by graphing; but it checks fairly easily.)

 Nov 19, 2014
 #2
avatar+95177 
+10
Best Answer

$$\\2^{(x-2)}-4^x+8^x=0\\
2^x*2^{-2}-2^{2x}+2^{3x}=0\\
2^x*2^{-2}-(2^{x})^2+(2^{x})^3=0\\
2^x(2^{-2}-2^{x}+(2^{x})^2)=0\\
2^x \mbox{ cannot equal zero so}\\
2^{-2}-2^{x}+(2^{x})^2=0\\
$Let $ y=2^x\\
2^{-2}-y+y^2=0\\
y^2-y+\frac{1}{4}=0\\\\
y=\frac{1\pm\sqrt{1-1}}{2}=\frac{1}{2}\\
$therefore$\\
\frac{1}{2}=2^x\\
2^{-1}=2^x\\
x=-1$$

Melody Nov 20, 2014

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