Exponential Equation?

Question

3^(2x+2) + 3^(x+1) - 12 = 0

move the 12 to the other side and use In to the left and side and divide by In3, but after this stage i get stuck, can someone show me all the working out for this question please??

thanks bhavin

Answer 1

You've missed the trick:

3^(2x+2) = 3^(x+1)^2

That means if u = 3^(x+1)

u² + u - 12 = 0

(u + 4)(u - 3) = 0

u = -4
u = 3


3^(x+1) = -4 → impossible
3^(x+1) = 3 → x = 0

Try it out:

x=0→ f(x) = 3² + 3 - 12 = 9 + 3 - 12 = 0

Please note that the answer above over-complicates things by making a bad substitution choice. They expect you to factor:

9 y² + 3y - 12
instead of
u² + u - 12 = 0

Honestly, which one would you rather factor?

Answer 2

3^(2x+2) + 3^(x+1) - 12 = 0

Let y = 3^x
then
3^(2x+2) = 9 y^2
3^(x+1) = 3y
and the equation becomes
9 y^2 + 3y - 12 =0
(3y + 4) (3y - 3) = 0
y = -4/3 or y =1
3^x = -4/3 has no solution, so
y=1=3^x gives x = 0 as the only solution

Answer 3

Let y = x + 1
3^(2y) + 3^y - 12 = 0
Let u = 3^y
u² + u - 12 = 0
(u + 4) (u - 3) = 0
u = 3 is the only acceptable solution.
3^y = 3
y = 1
x + 1 = 1
x = 0

Answer 4

Let Y= 3^x+1
Y^2= 3 ^2x+2
y^2+y-12=0
(Y+4)(Y-3)
3,-4
3^x+1=-4 no
3^x+1= 3^1
x+1=1
x=0

Answer 5

Maybe missin' somethin..but as I recall from high sch..(2x+2)=2(x+1) an' 3^2(x+1)=9^(x+1)..so add the powers..> 12^(x+1)=12..> x+1=1..x=0..?!

Answer 6

3^(2x+2)+3^(x+1)-12=0
3^2(x+1)+3^(x+1)-12=0
Substitute 3^(x+1) by a
a^2+a-12=0
(a+4)(a-3)=0
a= -4 or 3
If a= -4,then
3^(x+1)= -4.solution impossible
if a=3
3^(x+1)=3^0
or x+1=0
x= -1