# how to solve the equation 2*3^x + 3^(1-x)=5?

hala718 | Certified Educator

2*3^x + 3^(1-x) = 5

Let us expand the equation:

2*3^x + 3^1 * 3^-x = 5

2*3^x + 3/3^x =5

Let 3^x =y

2y +3/y = 5

2y^2 + 3 = 5y

2y^2 -5y + 3 = 0

Factorize:

(2y-3)(y-1) =0

y1= 3/2  ==> y1=3^x1=3/2  ==> x= log 3 (3/2)= log3 (3)-log 3 (2) = 1-log2

y2= 1 ==? y1 =3^x2=1 ==> x2=0

neela | Student

To solve 2*3^x+3^(1-x) = 5.  Rewrite

2*3^x + 3^1/3^x  = 5.

2t+3/t -5 = 0, where t = 3^x

Multiply by t.

2t^2-5t +3= 0.

2t^2 -3t -2t+3 = 0

t(2t-3)-1(2t-3) = 0

(2t-3)(t-1) =0.

2t-3 = 0, t-1 = 0.

2t-3 = 0, 2t = 3, 3^x = 3 = 3^1,  x = 1.

t-1 = 0, t = 1. 3^x = 1 = 3^0. x = 0.

giorgiana1976 | Student

This is an exponential equation and we'll solve it by using substitution technique.

First, we'll write 3^(1-x) = 3*3^-x = 3/3^x

The equation will become:

2*3^x + 3/3^x = 5

We'll multiply the first term and the term from the right side by 3^x:

2*3^2x + 3 - 5*3^x = 0

We'll substitute 3^x = t

The equation will become:

2t^2 - 5t + 3 = 0

t1 = [5+sqrt(25-24)]/4

t1 = (5+1)/4

t1 = 6/4

t1 = 3/2

t2 = (5-1)/4

t2 = 1

But:

3^x = t1

3^x = 3/2

We'll use the logarithms:

lg (3^x) = lg (3/2)

We'll use the power property to the left side:

x lg3 = lg (3/2)

We'll use the quotient rule to the right side:

x lg3 = lg 3 - lg 2

We'll divide by lg 3:

x = 1 - lg 2/lg 3

We'll find out the other solution:

3^x = t2

3^x = 1

3^x = 3^0

x=0

The solutions of the equation are:

x = 1 - lg 2/lg 3 and x=0