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What are all solution of the equation 289^x-12*17^x+11=0 ?

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colorcolour | Student, College Freshman | eNoter

Posted May 11, 2011 at 4:31 PM via web

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What are all solution of the equation 289^x-12*17^x+11=0 ?

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giorgiana1976 | College Teacher | Valedictorian

Posted May 11, 2011 at 4:36 PM (Answer #1)

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Since 289 is a power of 17, we'll write:

289 = 17^2

We'll raise both sides by x;

289^x = 17^2x

We'll re-write the equation:

17^2x - 12*17^x + 11 = 0

We'll replace 17^x by  t:

t^2 - 12t + 11 = 0

t^2 - t - 11t + 11 = 0

t(t - 1) - 11(t-1) = 0

(t - 1)(t - 11) = 0

We'll cancel each factor and w'ell get:

t - 1 = 0 => t1 = 1

t - 11 = 0 => t2 = 11

17^x = t1 <=> 17^x  = 1 <=> 17^x = 17^0

Since the bases are matching, we'll apply one to one property:

x = 0

17^x = t2<=> 17^x  = 11 <=> ln 17^x = ln 11 => x = ln 11/ln 17

The complete set of solutions of the equation is {0 ; ln 11/ln 17}.

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