Question

Solve for x.

log2(7x + 1) = log2(2 - x)


a) 1/8

b) 8

c) 1/4

d) 4

Answer 1

Take both sides to be some power of 2.

\[2^{\log_2(x)}=x\]

Answer 2

It will get rid of both of your logarithms and leave you with a more normal looking algebra equality.

Answer 3

(7x + 1) = (2 - x)

Answer 4

how would you solve for x ?????

Answer 5

Well both sides are equal right? That's what the equals sign means. So you can do stuff that involves adding or subtracting stuff, as long as you do it to both sides.

Remember if you have 2x+3x that's like saying 2 apples plus 3 apples. It's 5 apples, right? So 2x+3x=5x. That's just an example so you get the idea. I'll help you through it, but try it out, it's not so bad.

Answer 6

where would you move the numbers ??...
(7x + 1) = (2 - x)
-1 -1
7x = 1-x
+x +x
7x^2= 1
x^2 = 1/7

Answer 7

x^2=x*x
2x=x+x

which one are you doing here? You're almost right.

Answer 8

what do you mean which one am i doing ?

Answer 9

you added x to both sides, but you multiplied it somehow to get x^2, which is wrong.

Answer 10

okay i see ...
7x+x=3 ??

Answer 11

Almost, you changed the other part now.

7x+x=1 is where you should be.

Now you just need to add up the x's. See, if you have 7 of something and 1 of that same thing, how many of that thing do you have?

Answer 12

7x+1=2-x
7x+1-2+x=0
8x-1=0
8x=1
x=1/8

Answer 13

thanks

Answer 14

yw