Question

Logarithm Problem!!!
Pls Help,
Medal and a hearty thanks involved!!

Answer 1

\[1. (x+1)^{\log(x+1)}=100(x+1) .Solve for x.\]

Answer 2

For 0<x<1,the value of \[\log(1+x)+\log(1+x^2)+\log(1+x^4)..............\]

Answer 3

till infinity is??

Answer 4

2 Questions!

Answer 5

@hartnn

Answer 6

take log on both sides of the equation, for 1st

Answer 7

\[(x+1)^{\log(x+1)}=100(x+1).\]
\[\log(x+1)^{\log(x+1)}=\log(100(x+1) )\]
\[\log(x+1)\log(x+1)=\log(100(x+1) )\]
\[\log(2x+2)=\log(100(x+1))\]
\[(2x+2)=(100(x+1))=>2x+2=100x+100=>x=?\]
Umm I am wrong somewhere dunno where

Answer 8

log A * log A = (log A)^2
not log 2A

Answer 9

lol

Answer 10

oops yeah thanks

Answer 11

\[\log(x+1)^{\log(x+1)}=\log(100)+\log(x+1)\]

Answer 12

\[(x+1)^2=(100(x+1) )\]
x=?

Answer 13

then??

Answer 14

use the rule for left,
log A^B = B log A

Answer 15

\[\log(x+1)\log(x+1)=2+\log(x+1)\]

Answer 16

then??

Answer 17

yes
now pretend that log (x+1) = y
so you have a quadratic in y
right ?

Answer 18

O_o

Answer 19

are you sure the log has base 10 ?

Answer 20

Oh yeah rt..

Answer 21

Idunno

Answer 22

like y^2 = 2+y
isn't it ?
can u solve this ?

Answer 23

do we need to take base 10 ...cant we take whatever base is there in the question??

Answer 24

what the base in the question ??
and i think its log to base 10 only.
go on with it

Answer 25

yes\...;)

Answer 26

I get it now.......So then solve for y and substitute back the values,eh??

Answer 27

solving this y^2 = 2+y
is easy, you can even eyeball it :)

Answer 28

yes

Answer 29

y=1 or -2

Answer 30

you reversed the signs :P

Answer 31

nope

Answer 32

-1 and 2
isn't it ?

Answer 33

y^2-y-2=0

Answer 34

yeah

Answer 35

Oopsie ..I meant y-2 and y+1

Answer 36

then you're correct
so
values of y are ?

Answer 37

gotcha

Answer 38

good,
don't forget to verify your answer, when u get it

Answer 39

I get it.........Thanks a lot @hartnn and @INeedHelpPlease?

Answer 40

now the second one???

Answer 41

getting it??

Answer 42

@skullpatrol @nincompoop @lncognlto @robtobey @shamil98

Answer 43

first apply log A + log B = log AB rule
then take
x+1 as (x+1)(x-1) /(x-1)
and go on combining terms

Answer 44

Hint:
\[\log(1+x)+\log(1+x^2)+\log(1+x^4)..............\times \frac{1-x}{1-x}\]
\[\log(1+x) \times \frac{1-x}{1-x}+\log(1+x^2) \times \frac{1-x}{1-x}+\log(1+x^4) \times \frac{1-x}{1-x}..............\]

Answer 45

using the formula (a+b)(a-b) = a^2-b^2

Answer 46

marja :/^

Answer 47

What does that mean??

Answer 48

Marja??

Answer 49

he wants me to die :P

Answer 50

yeah a mathematical term..just like log

Answer 51

@INeedHelpPlease?
I am sorry........ur equation is not getting displayed due to errors in the website!!

Answer 52

refresh

Answer 53

here, take a medal from me for your atempt and be happy forever :P :)

Answer 54

still nothing.-_-

Answer 55

did u apply log A + log B rule ?

Answer 56

\[\[\log(1+x)+\log(1+x^2)+\log(1+x^4)..............\times \frac{1-x}{1-x}\]\]

Answer 57

\[\log(1+x)(1+x^2)(1+x^4)..............\times \frac{1-x}{1-x}\]

Answer 58

now that is correct ^

Answer 59

this is what is displayed........do u mean this @INeedHelpPlease? ??

Answer 60

yes

Answer 61

are u sure??

Answer 62

only look at the response above my last comment

Answer 63

this one has options
log(1-x)
logx
-log(1-x)
-logx

Answer 64

c

Answer 65

how is it c??

Answer 66

LOOK at the attached file @cambrige

Answer 67

I saw it.........\[\log{(1-x)/(1-x)}\]

Answer 68

how to proceed after that??

Answer 69

wot ? no
(1+x) (1-x)/ (1-x) = (1-x^2)/ (1-x)

so,
(1-x^2)(1+x^2)/(1-x) = (1-x^4)/(1-x)

getting this ?

Answer 70

to make it simple , you can keep that 1-x in the denomonator aside,
(1+x) (1-x) (1+x^2) (..)..

= (1-x^2) (1+x^2) (..)..

=(1-x^4 )(1+x^4) ....

Answer 71

Ur attachment states that the answer to the series is (1-x)/(1-x).r8???

Answer 72

no.......
thats just the first step to solve :P
multiplying and dividing by 1-x

Answer 73

Frankly speaking.M really not getting u!!:(

Answer 74

okay ....now gtin