3^(2x)+3^(x)=20?

Question

How do I find x?

The answer is log(4)/log(3) but I don't know how to get it.

log(20)/3(log(3)) =0.90 This is the wrong answer.

Thank you everyone for your answers.

Could someone please explain:

How
(3^x)^2+3^x-20=0
becomes
x^2+x-20=0

I got to:
3^x(3^x+1)=20

and

(-3^(x)±sqrt((3^x)^2-4*(3^x)^2*-20))/(2*(3^x)^2)

using (-b±sqrt(b²-4ac))/2a


but now I'm stuck.

Answer 1

Let y = 3^x
y² + y - 20 = 0
(y + 5)(y - 4) = 0
y = - 5 , y = 4
3^x = - 5 , 3^x = 4
Accept 3^x = 4 and take log base 3 of both sides. (let log mean log to base 3)
x log 3 = log 4
x = log 4

Answer 2

There's a trick.

3^(2x) and 3^(x) have the same base.

The base is 3.

You can therefore consider this equation to be a polynomial equation.

Its basicly (3^(x))² + 3^(x) - 20 = 0

using the (-b±sqrt(b²-4ac))/2a solver

you find 3^(x)= 4 or 3^(x) = -5

From this you use the logarithmic identity

y = Logb(x) <=> b^y = x

Log3(4) = x = Log(4)/Log(3) = 1.2618595
Log3(-5) = x = Log(-5)/Log(3) = impossible because you can't have a negative logarithm. You therefore reject the hypothesis that x = -5 .

Verficiation:

3^(2(1.2618595)) + 3^(1.2618595) =
15.999999 + 3.9999999 =~ 19.999999999 ~ 20

There you have it.

Answer 3

that's in all probability the long thank you to try this issue (you need to attempt the quadratic equation in case you dare...) yet I went with what popped into my head first. i began out via multiplying that total equation via 4. That clears out the nasty fraction. then you definitely're left with 3x^2 + 32x + 20. Now locate 2 numbers that multiply to 240 (three times 20) and upload to 32 (the middle term). I used the field approach because of the fact that's great user-friendly and enables you to easily pull out person-friendly factors. i'm no longer able to truly draw that yet i wish they're nevertheless coaching that approach haha. So now you need to have 2 linear factors: (3x+20)(x+4). The zeros (or x values) are x = -20/3 and x = -4

Answer 4

Thanks for a very simple question. Follow my step below.

Firstly, given the equation:
3^(2x) + 3^(x) = 20

Then, move all value from the right hand side to the left hand side:
3^(2x) + 3^(x) - 20 = 0

Let say, 3^2 = u
So, 3^(2x) + 3^(x) - 20 = 0 is u^2 + u - 20 = 0

From the new equation, we factorize it:
u^2 + u - 20 = 0
(x - 4) (x + 5) = 0

First answer:
x - 4 = 0
x = 4

Second answer:
x + 5 = 0
x = -5

Therefore, the answer should be:
x = 4 or x = -5

I hope you got the best answer for your question. Have a nice day!

Answer 5

Greetings,

Let z = 3^x

then z^2 + z - 20 = 0

(z+5)(z-4) = 0

z = -5 or z = 4

3^x = 4 , we can't get a negative value for 3^x

take log of both sides and isolate x

x = ln4 /ln3

Regards

Answer 6

ubiquito is right on the money........guys you can't add exponents when adding bases!! so there's no way that 3^2x + 3^x will equal 3^3x

Answer 7

3^(2x)+3^(x)=20

. . .by remainder theorem, . value of x = 1.26186
x = 1.26186

Answer 8

i agree by the solution given by happyper....but the thing is if subsitutue the solution you get back into the question, funny thing is you get 10.0.... and not 20...

Answer 9

edit: [haha! sorry. >.< Forget what I said. That other dude is way smarter. You can't add exponents like that. :p]