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September 22nd, 2019, 03:09 AM  #1 
Newbie Joined: Sep 2019 From: SLO Posts: 2 Thanks: 0  exponent
Hi, I need some idea how to start with this 4^(x2)  17*2^(x4) + 1 = 0 something with 4 and 2s are catch?? Last edited by skipjack; September 22nd, 2019 at 03:46 AM. 
September 22nd, 2019, 03:51 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 21,035 Thanks: 2271 
The equation factorizes as $\left(4^22^{{\large x}4}  1\right)\left(2^{{\large x}4}  1\right) = 0$.

September 22nd, 2019, 04:54 AM  #3 
Newbie Joined: Sep 2019 From: SLO Posts: 2 Thanks: 0 
How you got that ?

September 22nd, 2019, 07:35 AM  #4 
Senior Member Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 
$4^{x2} = 2^{2x4} = 16\cdot 2^{2x8}$

September 22nd, 2019, 08:52 AM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 3,034 Thanks: 1621 
$4^{x2} = 2^{2x4} = \dfrac{2^{2x}}{2^4} = \dfrac{1}{16} \cdot (2^x)^2$ $17 \cdot 2^{x4} = 17 \cdot \dfrac{2^x}{2^4} = \dfrac{17}{16} \cdot 2^x$ Let $t=2^x$ ... $4^{x2} 17 \cdot 2^{x4} + 1=0 \implies \dfrac{t^2}{16}  \dfrac{17t}{16} + 1=0$ multiply every term by 16 ... $t^217t+16=0 \implies (t1)(t16)=0 \implies t=1 \text{ or } t=16$ finish it 
September 22nd, 2019, 12:20 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 21,035 Thanks: 2271  

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