4^2x - 2 * 4^(x+4) + 4^8 = 0
- Thread starter DemiGod
- Start date
Hello, DemiGod!
\(\displaystyle \text{We have: }\;4^{2x} - 2\!\cdot\!4^4\!\cdot\!4^x + 4^8 \:=\:0\)
\(\displaystyle \text{Factor: }\;\left(4^x - 4^4\right)^2 \:=\:0\)
\(\displaystyle \text{Therefore: }\;4^x - 4^4 \:=\:0\quad\Rightarrow\quad 4^x \:=\:4^4\quad\Rightarrow\quad\boxed{ x \:=\:4}\)
\(\displaystyle \text{Factor: }\;4(3x-2)^{-\frac{4}{3}}\,\bigg[-x + (3x - 2)\bigg] \:=\:0 \quad\Rightarrow\quad 4(3x-2)^{-\frac{4}{3}}(2x-2) \:=\:0\)
\(\displaystyle \text{Then: }\;(3x-2)^{-\frac{4}{3}} \:=\:0\quad\hdots\;\text{ not possible}\)
\(\displaystyle \text{And: }\;\;2x-2 \:=\:0 \quad\Rightarrow\quad\boxed{x \:=\:1}\)
\(\displaystyle \text{We have: }\;4^{2x} - 2\!\cdot\!4^4\!\cdot\!4^x + 4^8 \:=\:0\)
\(\displaystyle \text{Factor: }\;\left(4^x - 4^4\right)^2 \:=\:0\)
\(\displaystyle \text{Therefore: }\;4^x - 4^4 \:=\:0\quad\Rightarrow\quad 4^x \:=\:4^4\quad\Rightarrow\quad\boxed{ x \:=\:4}\)
\(\displaystyle \text{Factor: }\;4(3x-2)^{-\frac{4}{3}}\,\bigg[-x + (3x - 2)\bigg] \:=\:0 \quad\Rightarrow\quad 4(3x-2)^{-\frac{4}{3}}(2x-2) \:=\:0\)
\(\displaystyle \text{Then: }\;(3x-2)^{-\frac{4}{3}} \:=\:0\quad\hdots\;\text{ not possible}\)
\(\displaystyle \text{And: }\;\;2x-2 \:=\:0 \quad\Rightarrow\quad\boxed{x \:=\:1}\)
Any chance you can explain the second problem a bit?
I don't understand exactly how you factored that into each next step. After the factor, i get it, but the whole thing in []'s leaves me clueless. What happened to the exponent? And the second 4? I know they're gone, but what did you divide by to get rid of them, etc.
- D
I don't understand exactly how you factored that into each next step. After the factor, i get it, but the whole thing in []'s leaves me clueless. What happened to the exponent? And the second 4? I know they're gone, but what did you divide by to get rid of them, etc.
- D
I also have a bit of a hard time following Soroban's (he's a former teacher, you know!!)
Maybe you'll follow this:
-4x(3x - 2)^(-4/3) + 4(3x - 2)^(-1/3) = 0 ; re-arrange:
4x(3x - 2)^(-4/3) = 4(3x - 2)^(-1/3) ; divide by 4:
x(3x - 2)^(-4/3) = (3x - 2)^(-1/3) ; re-arrange:
x(3x - 2)^(-1)(3x - 2)^(-1/3) = (3x - 2)^(-1/3) ; divide by (3x - 2)^(-1/3):
x(3x - 2)^(-1) = 1 ; re-arrange:
x / (3x - 2) = 1
3x - 2 = x
2x = 2
x = 1
Maybe you'll follow this:
-4x(3x - 2)^(-4/3) + 4(3x - 2)^(-1/3) = 0 ; re-arrange:
4x(3x - 2)^(-4/3) = 4(3x - 2)^(-1/3) ; divide by 4:
x(3x - 2)^(-4/3) = (3x - 2)^(-1/3) ; re-arrange:
x(3x - 2)^(-1)(3x - 2)^(-1/3) = (3x - 2)^(-1/3) ; divide by (3x - 2)^(-1/3):
x(3x - 2)^(-1) = 1 ; re-arrange:
x / (3x - 2) = 1
3x - 2 = x
2x = 2
x = 1