# What is x if 13^x - 20=13^(3-x) ?

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to solve 13^x - 20 = 13^ (3 - x)

13^x - 20 = 13^ (3 - x)

=> 13^x - 20 = 13^3/ 13^x

let 13^x = y

=> y - 20 = 13^3 / y

=> y^2 - 20 y - 13^3 = 0

y1 = [20 + sqrt (20^2 + 4*13^3)]/ 2

= 10 + sqrt 2297

y2 = 10 - sqrt 2297

As 13^x = y

13^x = 10 + sqrt 2297

x log 13 = log (10 + sqrt 2297)

x = log (10 + sqrt 2297) / log 13

=> x = 1.5825 (approx)

Therefore x = log (10 + sqrt 2297) / log 13

neela | High School Teacher | (Level 3) Valedictorian

Posted on

What is x if 13^x - 20=13^(3-x).

13^x - 13^(3-x) - 20 = 0.

We multiply by 13^x :

(13^x)^2 - 20*13^x - 13 ^3 = 0.

y^2 - 20y - 2197 .

y1 = {20+ sqrt(20^2+4*2197)}/2 = 57.9270

y2 = {20-sqrt(20^2+4*2197)}/2 which is negative.

Y1 = 57.9270 gives  13^x = 57.92703.

=>  x= log57.92703)/log13.

x = 1.5826 nearly.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll remember the quotient property of exponentials:

13^(3-x) = 13^3/13^x

We'll re-write the equation:

13^x - 20 - 13^3/13^x = 0

We'll multiply by 13^x both sides:

13^2x - 20*13^x - 13^3 = 0

We'll substitute 13^x = t:

t^2 - 20t - 2197 = 0

t1 = [20+sqrt(400 + 8788)]/2

t1 = (20 + sqrt9188)/2

t1 = 10+sqrt2297

t2 = 10-sqrt2297

13^x = 10+sqrt2297

log 13^x = ln (10+sqrt2297)

x = log (10+sqrt2297)/ln 13

x = 1.5825 approx.

13^x = 10 - sqrt2297 impossible! ( since 10 < sqrt2297 => 13^x = negative value, that is impossible).

The equation has only one solution, x = 1.5825 approx.