Question

how do i solve (2^x-3)^2+2=9

Answer 1

idk tell me the answer

Answer 2

1. subtract 2 on both sides
2. then take square root on both sides.
3. add 3 to both sides.
4. take log on both sides

Answer 3

log(3+/-sqrt(7),2)

Answer 4

how do i take the square root on both sides. would that be? Square root of 2^x-3= square root of 7

Answer 5

that is right.

Answer 6

then i have square root of 2^x = square root of 7+3, if that is correct. how do i do the log

Answer 7

no. you have
\[2^x= \sqrt 7 +3\]

Answer 8

log(2^x) = log(sqrt(7) + 3)

x = log(sqrt(6) + 3)/log(2)

Answer 9

Okay, i see. now how do i explain the log

Answer 10

what do you mean?

logging something is the opposite of exponentiating something

Answer 11

Yes, i know that much. I'm sorry, let me rephrase that. What are the steps you took to get that answer for log. I can't just put an answer because that does me no good.

Answer 12

so far i have log(2^x) =log (root(7)+3)

Answer 13

log a^b = b log a
so log 2^x = x log 2 = log(sqrt7 +3)
x = log(sqrt7 +3)/log 2

Answer 14

Honestly, I am lost after x log 2. I keep getting an end result of x=log2 (sq root(7) +3). Can you try and explain to me why I am wrong. Maybe that might help

Answer 15

(2^x\[(2^x - 3 )^2 = 7 \rightarrow 2^{2x} - 3 \times 2^{x+1} + 9 = 7\]

let \[2^x = y\]
so \[y^2 - 3 \times 2 \times y + 2 = 0\] -> Solving....

y` = 3 + sqrt{7} `n y``= 3 - sqrt{7}
back to 2^x = y ....

\[x = \log_{2} {3\pm \sqrt{7}}\]