Solve for x over the real numbers:
-4-2^(3 x-4)-2^(3 x-3)+2^(3 x-2)=0
The left hand side factors into a product with three terms:
1/16 (2^x-4) (16+2^(2 x)+2^(x+2))=0
Multiply both sides by 16:
(2^x-4) (16+2^(2 x)+2^(x+2))=0
Split into two equations:
2^x-4=0 or 16+2^(2 x)+2^(x+2)=0
Add 4 to both sides:
2^x=4 or 16+2^(2 x)+2^(x+2)=0
4=2^2:
2^x=2^2 or 16+2^(2 x)+2^(x+2)=0
Equate exponents of 2 on both sides:
x=2 or 16+2^(2 x)+2^(x+2)=0
Simplify and substitute y=2^x:
16+2^(2 x)+2^(x+2) = 16+4×2^x+(2^x)^2 = y^2+4 y+16 =0:
x=2 or y^2+4 y+16=0
Subtract 16 from both sides:
x=2 or y^2+4 y=-16
Add 4 to both sides:
x=2 or y^2+4 y+4=-12
Write the left hand side as a square:
x=2 or (y+2)^2=-12
(y+2)^2=-12 has no solution since for all y on the real line, (y+2)^2 ≥0 and -12<0:
Answer: | x=2
+23251 Solve: 23x-2 - 23x-3 - 23x-4 - 4 = 0
23x-2 = 23x · 2-2 = 23x · (1/4) = 23x / 4
Similarly,
23x-3 = 23x / 8
23x-4 = 23x / 16
So: 23x-2 - 23x-3 - 23x-4 - 4 = 0 ---> 23x / 4 - 23x / 8 - 23x / 16 - 4 = 0
Mulitply both sides by 16: ---> 16 [ 23x / 4 - 23x / 8 - 23x / 16 - 4 ] = 16 [ 0 ]
---> 4 · 23x - 2 · 23x - 1 · 23x - 64 = 0
Add 64 to both sides: ---> 4 · 23x - 8 · 23x - 1 · 23x = 64
Factor out 23x: ---> 23x [ 4 - 2 - 1 ] = 64
Simplify: ---> 23x [ 1 ] = 64
Simplify: ---> 23x = 64
Rewrite 64: ---> 23x = 26
---> 3x = 6
---> x = 2
(But, this isn't a quadratic equation!)
+26388 2^(3x-2)-2^(3x-3)-2^(3x-4)-4=0
\(\begin{array}{|rcll|} \hline 2^{3x-2}-2^{3x-3}-2^{3x-4}-4 &=&0 \\\\ 2^{3x-2}-2^{3x-2-1}-2^{3x-2-2}-4 &=&0 \\ 2^{3x-2}\cdot (1 -2^{-1}-2^{-2} ) -4 &=&0 \\ 2^{3x-2}\cdot (1 -\frac12-\frac14 ) -4 &=&0 \\ 2^{3x-2}\cdot (\frac14 ) -4 &=&0 \qquad &|\qquad +4\\ 2^{3x-2}\cdot (\frac14 ) &=& 4 \qquad &|\qquad \cdot 4\\ 2^{3x-2} &=& 4\cdot 4\qquad &|\qquad 4\cdot 4 = 16 = 2^4\\ 2^{3x-2} &=& 2^4\\ 3x-2 &=& 4 \qquad &|\qquad +2 \\ 3x &=& 4+2 \\ 3x &=& 6 \qquad &|\qquad :3 \\ x &=& \frac63 \\ \mathbf{x} & \mathbf{=} & \mathbf{2} \\ \hline \end{array} \)
