My Math Forum exponent

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September 22nd, 2019, 03:09 AM   #1
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exponent

Hi,

4^(x-2) - 17*2^(x-4) + 1 = 0

something with 4 and 2s are catch??
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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 $\left(4^22^{{\large x}-4} - 1\right)\left(2^{{\large x}-4} - 1\right) = 0$. Thanks from topsquark, DarnItJimImAnEngineer and Medo73
 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 $4^{x-2} = 2^{2x-4} = 16\cdot 2^{2x-8}$ Thanks from Medo73
 September 22nd, 2019, 08:52 AM #5 Math Team     Joined: Jul 2011 From: Texas Posts: 3,034 Thanks: 1621 $4^{x-2} = 2^{2x-4} = \dfrac{2^{2x}}{2^4} = \dfrac{1}{16} \cdot (2^x)^2$ $17 \cdot 2^{x-4} = 17 \cdot \dfrac{2^x}{2^4} = \dfrac{17}{16} \cdot 2^x$ Let $t=2^x$ ... $4^{x-2} -17 \cdot 2^{x-4} + 1=0 \implies \dfrac{t^2}{16} - \dfrac{17t}{16} + 1=0$ multiply every term by 16 ... $t^2-17t+16=0 \implies (t-1)(t-16)=0 \implies t=1 \text{ or } t=16$ finish it Thanks from topsquark, DarnItJimImAnEngineer and Medo73
September 22nd, 2019, 12:20 PM   #6
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Quote:
 Originally Posted by Medo73 How you got that ?
As $4^{{\large x} - 2} = 4^24^{{\large x} - 4} = 4^2\left(2^{{\large x} - 4}\right)^2$,
the original equation is equivalent to $4^2u^2 - 17u + 1 = 0$, where $u = 2^{{\large x} - 4}$,
which is a quadratic equation that factorizes easily.

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