Question

Help me with guide.
The decimal part of a real number a is denoted by {a}. Find the smallest real number x such that x>5 and {log(x+2)}+{log(x)}=1

Answer 1

@mirandabell902 please help

Answer 2

@StudyGurl14 please help, trollwillrule is my mentor. he gave me this qn

Answer 3

sorry, I don't know logs very well

Answer 4

{log(x+2)}+{log(x)}=1
you know 0<={}<1
so
0<={log(x+2)}<1
0<= {log(x)}<1
find them with this condition
for example {log(x+2)}=.6 {log(x)}=.4
or {log(x+2)}=.1 or {log(x)}=0.9

Answer 5

@zh123456

Answer 6

ok i draw the answer and also try this qn

Answer 7

dw:1387607283717:dw

Answer 8

nvm i sholw u the answer somewhere when we meet tmr

Answer 9

A polynomial fx can be expressed in the form fx = [Ax2 plus Bx] times Dx plus 2x -5, where Dx is a divisor. If D1 = D2 = 1 and x-1 and x-2 are factors of fx, find f3.

Answer 10

\[\lfloor \log(x+2) \rfloor+\lfloor \log(x)\rfloor+\left\{ \log(x+2) \right\}+\left\{ \log(x) \right\}=K \] \[\lfloor \log(x+2) \rfloor+\lfloor \log(x) \rfloor =K-1\] \[\log(x^2+2x) + \log(10)=K \] Now do it. People here cannot even solve this lol

Answer 11

this is a huge hint alr