Verify if exponential equation has any roots. 4^x=20-4^(3-x)

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

4^x = 20 - 4^(3-x)

We need to find the values of x, if any, that satisfies the exponential equation.

First we will rewrite 4^(3-x) = 4^3/4^x

==> 4^x = 20 - 4^3/4^x

Now we will assume that :

y= 4^x

==> y = 20 -  64/y

Now we will multiply by y.

==> y^2 = 20y - 64

==> y^2 - 20 + 64 = 0

Now we will factor.

==> (y-16)(y-4) = 0

==> y1= 16 ==> 4^x = 16 = 4^2 ==> x1 = 2

==> y2= 4 ==> 4^x = 4 = 4^1 ==> x2= 1

Then we have two values for x that satisfies the equation:

x = { 1, 2}

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll re-write the term 4^(3-x) recalling the quotient property of exponentials:

4^(3-x) = 4^3/4^x

We'll re-write the equation:

4^3/4^x + 4^x - 20 = 0

We'll multiply by 4^x both sides:

4^3 + 4^2x - 20*4^x = 0

We'll substitute 4^x = t:

t^2 - 20t + 64 = 0

We'll apply quadratic rule:

t1 = [20+sqrt(400 - 256)]/2

t1 = (20+sqrt144)/2

t1 = (20+12)/2

t1 = 16

t2 = (20-12)/2

t2 = 4

But t1 = 4^x => 16 = 4^x

We'll put 16 = 4^2

4^x = 4^2

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

x = 2

t2 = 4 => 4^x = 4 => x = 1

The equation has 2 real solutions and they are: {1 ; 2}.

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