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 Algebra Pre-Algebra and Basic Algebra Math Forum

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

Hi,

I need some idea how to start with this

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

something with 4 and 2s are catch??
Attached Images mat17a.jpg (15.8 KB, 0 views)

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. Tags exponent Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post hhmathgeek Algebra 13 January 21st, 2016 07:16 PM The_Ys_Guy Algebra 3 May 1st, 2015 07:21 PM bml1105 Algebra 12 July 27th, 2014 11:40 AM confucius Algebra 1 February 18th, 2014 08:15 PM Thinkhigh Calculus 3 March 2nd, 2012 06:42 AM

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