Expert Answers
hala718 eNotes educator| Certified Educator

169^x + 13^x = 79

Let us move 79 to the left side:

==> 169^x + 13^x - 79 = 0

Now let us rewrite 169 = 13^2

==> (13^2)^x + 13^x - 79 = 0

==> (13^x)^2 + 13x - 79 = 0

Let y = 13^x :

==> y^2 + y - 79 = 0

==> y1= [-1+sqrt(1+4*79)]/2 = [ -1 + 17]/2 = 8

==> y2= [ -1 -17]/2 = -9

But y= 13^x

==> 8 = 13^x1 ==> x1= log 13 (8)

==> -9 = 13^x2 ==> inmpossible>

Then the only solution is :

x= log 13 (8)

william1941 | Student

We have to solve for x from the equation given which is : 169^x + 13^x = 79

Now 169 = 13^2

So, 169^x + 13^x = 79

=> 13^2^x + 13^x =79

=> 13^x^2 + 13^x = 79

Take 13^x as R

=> R^2 + R = 79

=> R^2 + R -79 = 0

Solving the equation, the roots we get are:

R1= [-1 + sqrt(1^2+4*79))] / 2

= ( -1 + sqrt 317 ) / 2

R2 = [-1 - sqrt(1^2+4*79))] / 2

= ( -1 - sqrt 317 ) / 2

Now R1 = 13^X = ( -1 + sqrt 317 ) / 2

So taking log on both the sides

X= log [( -1 + sqrt 317 ) / 2] / log 13

We are not considering R2 as it is a negative root and the logarithm for negative numbers is not defined.

Therefore x is log [( -1 + sqrt 317 ) / 2] / log 13

 

neela | Student

169^x +13^x = 79.

Put 13^x = y, then 169^x= (13^x)^.

Thus the given equation is :

y^2+y = 79

y^2+y -79 = 0

y = (-1+sqrt(1^2+4*79))/2

y = (-1+sqrt317)/2 or

y = (1-sqrt317)/2 which we ignore as it gives imaginary results.

13 ^x =  (-1+sqrt317)/2

 x log13 =  (-1+sqrt317)/2.

x = log(-1+sqrt317)/2 )/log13

x = 0.829840617

giorgiana1976 | Student

We'll use the substitution method to solve the exponential equation.

13^x=t

169^x=(13^2)^x=(13^x)^2=t^2

t^2 + t  = 79

We'll subtract 79 both sides:

t^2 + t - 79 = 0

We'll apply the quadratic formula:

t1 = [-1+sqrt(1+316)]/2=16/2=approx. 8

13^x=t1

13^x=8 

We'll take logarithms both sides:

 log 13^x = log 8

We'll use the power property of logarithms:

x log 13 = log 8

We'll divide by log 13:

x = log 8/log 13

t2=(-1-17)/2=approx. -9

13^x>0 so it's impossible that 13^x=-9!

The only solution is x=log 8/log 13.

Access hundreds of thousands of answers with a free trial.

Start Free Trial
Ask a Question