Question

solve Log_2(x-5)=3-log_2(x-3)

Answer 1

\[\log_{2}(x-3)=3-\log_{2}(x-5) \]

Answer 2

Exponentiate to base 2 for both sides and simplify using exponent/logarithm rules.\[\huge 2^{\log_2(x-3)}=2^{3-\log_2(x-5)}\]

Answer 3

that's completely wrong thanks though

Answer 4

thats not wrong, it will give the correct answer. there are other ways to do it, maybe even easier, but what is written is not wrong.

Answer 5

well according to my math book it is

Answer 6

there are many ways to do this problem. your math book isnt saying its wrong, its just not showing you this method.

Answer 7

are u going to explain?

Answer 8

Your math book is probably showing you something like this:\[\log_2(x-3)=3-\log_2(x-5)\Longrightarrow \log_2(x-3)+\log_2(x-5)=3\]\[\Longrightarrow \log_2((x-3)(x-5))=3\]\[\Longrightarrow (x-3)(x-5)=2^3\Longrightarrow x^2-8x+15=8\]\[x^2-8x+7=0\Longrightarrow (x-7)(x-1)=0\Longrightarrow x=7,x=1\]We cant use 1 however, because then we would be taking logs of negative numbers. Thus x=7 is the only solution.

Answer 9

ok here is my problem, that is exactly what i did but i did not square the 3 and therefor my answer was 6 not 7 why are u squaring the 3

Answer 10

With yak's method, we would get:\[2^{\log_2(x-3)}=2^{3-\log_2(x-5)}\Longrightarrow x-3=\frac{2^3}{(x-5)}\Longrightarrow (x-3)(x-5)=2^3\]everything is the same from there.

Answer 11

are you talking about the:\[2^3\]?That comes from the log property:\[\log_bx=y\Longrightarrow b^y=x\]

Answer 12

your method id the one i was taught, i understand what is going on now, sorry for being impatient im just frustrated, iv been doing this problem for an hour because im an idiot
and i have a test Thursday . thank you though

Answer 13

I was on the right track i just had x^2-8x+12 because i didn't make it 2^3 therefor my factoring was off. thank you for sparing me another hour of frustration though