solve (2^2x)-3 * (2^x+1) + 8 = 0, (3^2x+1) - 10 * (3^x) + 3

[jackayyy]

I am in desperate need of help on my advanced math project. I need to solve these two problems, but I can't figure them out.

1) (2^2x)-3 * (2^x+1) + 8 = 0

2) (3^2x+1) - 10 * (3^x) + 3 = 0

Thanks so much! :]

 

[stapel]

1) (2^2x)-3 * (2^x+1) + 8 = 0

2) (3^2x+1) - 10 * (3^x) + 3 = 0

Your formatting is ambiguous.

1) Do you mean any of the following?

. . . . .[(2²x) - 3][2^x + 1] + 8 = 0

. . . . .2²x - 3[2^x + 1] + 8 = 0

. . . . .2²x - 3(2^{x + 1}) + 8 = 0

. . . . .2^2x - 3(2^{x + 1}) + 8 = 0

Or did you mean something else?

2) Same question.

When you reply, please show everything you have tried so far. Thank you.

Eliz.

 

[jackayyy]

they are listed like this in my book;

1. 2^2x - 3 * 2^{x + 1} + 8 = 0
2. 3^2x+1 - 10 * 3^x + 3 = 0

the book doesnt give any grouping symbols, so i don't know where to start, or how to really read the problem. where i put the asterik that means the book just gives a multiplication dot. theres no parenthesis to indicate wheather or not the - 3(first problem) belongs with 2^2x or 2^{x + 1}. same with the second problem i don't know weather the - 10 belongs with 3^2x+1 or 3^x . sorry if that doesn't make sense.


well i have been working on the 2nd problem first because my book lists the answer as 1 or -1. and i figured if i could figure out how to solve the problem i could apply those concepts to the first problem.

work


(3^2x+1 - 10) * (3^x + 3 )= 0
i grouped those together and then tried the foil method to get this

9 ^3x+1 - 30^x + 9^2x+1 - 30 = 0

then i tried factoring by grouping to get this

9^2x+1(1^x) - 30(1^x)=0
9^4x - 30^x= 0

after i did this i then tried to substitute the answers in to see if i was on the right track, and then answers didn't work, so this approach was totally wrong.

i tried then to multiply the -10 by the 3^x

to get 3^2x+1-30^x + 3 = 0

substituting the answer into that works, but i have no idea where to go from there.
i hope that helps :]

 

[stapel]

they are listed like this in my book;

1. 2^2x - 3 * 2^{x + 1} + 8 = 0
2. 3^2x+1 - 10 * 3^x + 3 = 0

In what follows, I will start with my guess as to the book's meaning.

1) 2^2x - 3(2^{x + 1}) + 8 = 0

Since 2^{x + 1} = (2^x)(2¹) = 2(2^x), this equation simplifies to be a quadratic in 2^x:

. . . . .[2^x]² - 6[2^x] + 8 = 0

Factor, or apply the Quadratic Formula (with a = 1, b = -6, c = 8), to find the values of 2^x. Then solve the resulting exponential equations.

2) 3^{2x + 1} - 10(3^x) + 3 = 0

Since 3^{2x + 1} = (3^2x)(3¹) = 3(3^2x), this is just another quadratic, this time in 3^x:

. . . . .3[3^x]² - 10[3^x] + 3 = 0

Factor, or apply the Quadratic Formula (with a = 3, b = -10, c = 3), to find the values of 3^x. Then solve the resulting exponential equations.

Eliz.

 

[jackayyy]

oh wow. how did i not see that lol. i was thinking okay maybe i need to use logs? but seriosuly thank you so much. you just saved my math grade :]