Question

There are two functions g(x) and f(x), which intersect at point A. I need to find an oordinate of that point. Functions:
\[f(x)=5-\log _{2}(x-4)\]
\[g(x)=\log _{2}x\]
How I do:
\[5-\log _{2}(x-4)=\log _{2}x\]
\[-\log _{2}(x-4)-\log _{2}x=-5|:(-1)\]
\[\log _{2}(\frac{ x+4 }{ x })=5\]
\[\log _{2}(\frac{ x+4 }{ x })=\log _{2}32\]
So I make a system of equation:
\[\frac{ x+4 }{ x }=32\]
\[\frac{ x+4 }{ x }>0\]
\[x \neq 0\]
I solve it and get that x=28. When put x=28 into functions, but get wrong answer. can anyone help me?: ) The answer should be 2.

Answer 1

@satellite73

Answer 2

@amistre64

Answer 3

@phi

Answer 4

@anybody?:o

Answer 5

gonna have to take a break halfway thru .. thats alot of reading

Answer 6

f(x)=5-log2(x-4)
g(x)=log2x

is that a log base 2?

Answer 7

oh, maybe you could do it by yourself and tell me if you get 2?

Answer 8

yes, the base is 2. I used Latex, doesnt it work?

Answer 9

oh, you did dint you :) its been a long day

Answer 10

\[f(x)=5-log_2(x-4)\\
g(x)=log_2(x)\]

equate them to find out where they are the same, so a system of equations is fine

\[5-log_2(x-4)=log_2(x)\]
\[5=log_2(x)+log_2(x-4)\]
\[5=log_2(x(x-4))\]
\[2^5=2^{log_2(x(x-4))}\]
\[2^5=x(x-4)\]
\[32=x^2-4x\]
\[36=x^2-4x+4\]
\[36=(x-2)^2\]
\[\pm6=x-2\]
\[2\pm6=x=8,-4\]
isolate for any ill gotten values

Answer 11

oh, right. it was +,so I needed to multiply. Thanks a lot!

Answer 12

id go with x=8 meself .... since log(-4) is not real .....

Answer 13

yeah, but I need to find y(oordinate), so I will get 2, because f(2)=5-log(base 2)8=2

Answer 14

thank you alot:)

Answer 15

log2(8)

log2(2^3)

3 log2(2) = 3

Answer 16

well, there is 5 in the function too;) so 5-3=2 : ) You should get some rest! THANKS!

Answer 17

x = 8, so find f(8) and g(8)

Answer 18

lol, fine fine fine ;) good luck

Answer 19

good luck you too!

Answer 20

log2(8-4) = log(2^2) = 2

5 - 2 = 3

Answer 21

oh haha, my mistake!

Answer 22

maybe the given answer is wrong