Question

4log2(x^2+2x+1)=16

Answer 1

\[\Large4\log_2(x^2+2x+1)=16\]

Answer 2

Factorize \(\Large x^2+2x+1\) first :)
(Hint: it's a perfect square)

Answer 3

\[Log_ab=c~~~~~~~~~~~a^c=b\]I was thinking of this.

Answer 4

(x + 1 - √2)(x + 1 + √2)

Answer 5

\[ALSO:~~~~alog(b)=Log(b^a)\]

Answer 6

there is an easier way to do this.

Answer 7

@pinknabastak what you answered is \(x^2+2x-1\) not \(x^2+2x+1\)

Answer 8

@pinknabastak and I said it's a perfect square :)

Answer 9

\[4Log_2(x^2+2x+1)=16\]divide both side by 4\[Log_2(x^2+2x+1)=4\]

Answer 10

\[Log_ab=c~~~~~~~~~a^c=b\]

Answer 11

ok @SolomonZelman soo what would be next?

Answer 12

@SolomonZelman you don't call that an "easier" way. you call that a harder way

Answer 13

\[x^2+2x+1=2^4\]

Answer 14

2^4=(x^2+2x+1)

Answer 15

oh yea haha

Answer 16

\[x^2+2x+1=2^4\]

Answer 17

@pinknabastak
\[4\log_2(x^2+2x+1)=16\\4\log_2(x+1)^2=16\\8\log_2(x+1)=16\\\log_2(x+1)=2\\x+1=2^2\\x=3\]

Answer 18

whats the next step in your way @SolomonZelman ? because thats the way my teacher has explained it before but still didnt exactly click

but @kc_kennylau i see how your way is fast and it answers it as 3 but i would like to see all ways

Answer 19

recalling.
\[Log_2(x^2+2x+1)=4\]

Answer 20

yes so how does that get to x=3?

Answer 21

\[Log_2(x^2+2x+1)=4Log_22\]\[Log_2(x^2+2x+1)=Log_2(2^4)\]\[Log_2(x^2+2x+1)=Log_216\]\[x^2+2x+1=16\]

Answer 22

and from then you can do it....

Answer 23

But either way, I like how @kc_kennylau does it

Answer 24

I like your way too lolz :)

Answer 25

The question of this type is just asking

"Do you know the identities and how to use them?"

Answer 26

ok thanks...:)