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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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}.

Educator Approved

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

The equation 289^x-12*17^x+11=0 has to be solved. Notice that 289 = 17^2. The equation can be rewritten as a quadratic equation as follows:

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

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

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

(17^x)^2 - 11*17^x - 17^x + 11 = 0

(17^x)(17^x - 11) - 1(17^x - 11) = 0

(17^x - 1)(17^x - 11) = 0

For 17^x - 1 = 0

17^x = 1 or x = 0

17^x - 11 = 0 gives 17^x = 11 or x = log 11/log 17

The solution of the equation is x = 0 and x = (log 11)/(log 17)

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