Question

What is the extraneous solution found in solving the equation logsmall2 4x + logsmall2 (x + 1) = 3
A) -2
B) -1
C) 1
D) 2
E) -3

Answer 1

sub it into your equation, it will be the one that is false

Answer 2

how would it be the one that is false? Don't you want to find the solution?

Answer 3

when you solve it, you will find that x will give you that as a solution, but when you test it it does not work
extraneous
EXTRA solution that does not work
ever hear of a false positive on an exam
You took an exam -say pregnancy and it came out positive
but you go to the doctor and they test your blood and find out you were never pregnant
the exam gave you a false positive
same thing in math except we call it extraneous

Answer 4

I'm not getting it it's all coming out the wrong answer

Answer 5

the first step to answering this question is to solve for x in
\[ log_2(4x) + log_2(x+1) = 3\]

to solve, I would begin by using
log(a)+log(b) = log(a*b)
to rewrite the equation
then make both sides powers to the base 2 to get rid of the log

Answer 6

\[\log_{2}{4x}+\log_{2}{x+1} \]
=>\[\log_{2}{4x(x+1)} \]
=>\[\log_{2}{4x^2+4} \]

Answer 7

where is x next 4 ?

Answer 8

Oops!!! sorry it's 4x.

Answer 9

rewrite it please

Answer 10

\[\log_{2}{4x}+\log_{2}{x+1} \]
=>\[\log_{2}{4x(x+1)} \]
=>\[\log_{2}{4x^2+4x} \]

Answer 11

zac, are you following this?

Answer 12

yes and no

Answer 13

Then, it will be
\[\log_{2}{4x^2+4x}=3 \]
=>\[2^3=4x^2+4x\]
Now, solve it & gt ur answer:)

Answer 14

hint you will need to use the quadratic formula to solve for x

Answer 15

Yeah, precal is correct.

Answer 16

so to summarize, we started with your equation
\[ log_2(4x) + log_2(x+1) =3 \]
we use a (very important) property of logs: we can rewrite the sum of 2 logs (of the same base) as the log of products:
that means we get
\[ log_2(4x(x+1))=3 \]
now make both sides the exponents to base 2:
\[ 2^{log_2(4x(x+1))}=2^3 \]
hopefully you know the stuff on the left "undoes" the log
\[ 4x(x-1)= 8 \]
can you solve for x at this point?

Answer 17

use \[-b \pm \sqrt{b^2-4ac}\over2a\]