If log4 + log(17x+1) = log(3x+7) - 0^(0!) + 0^(0) + 9 Mod 11 - 243 Mod 13 , then value of x is ? No

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05 Apr 2017 Simplification

If log4 + log(17x+1) = log(3x+7) - 0^(0!) + 0^(0) + 9 Mod 11 - 243 Mod 13 , then value of x is ?
Note:
Assume all log represented above is to the base of 10.
^ denotes power of
! denotes factorial

OPtion

1) 33/19
2) 66/19
3) 66/18
4) 33/18
5) 68/38
6) 74/38
7) 64/38
8) 32/17
9) 64/36
10) None of these

Solution

Considering base is 10, so 1 could be represented as Log10, 2 as Log 100…..
Let’s revise the formula for Logs
Log(XY) = LogX+LogY
Applying the formula we get:

= > log4 + log(17x+1) = log(3x+7) - 0^(0!) + 0^(0) + 9 Mod 11 - 243 Mod 13
= > log(68x + 4) = log(3x+7) - 0 + 1 + 9 - 9
= > log(68x + 4) = log(3x+7) + 1
= > log(68x + 4) = log(3x+7) + log 10
= > log(68x + 4) = log(30x+70)
= > log(68x + 4) = log(30x+70)
= > 68x + 4 = 30x+70
= > 38x = 66
= > 19x = 33
= > x = 33/19
Option 1)

Correct Option: 1 Best Solution (3)

RASHID   7 years AGO

log4+log(7x+1)=log(3x+7)-0^(0!)+0^(0)+9mod11-243mod13

=>log(4*(17x+1))=log(3x+7)-0+1+9-9 ( '.' 0^(0)=1; & 0^(0!)=1;)

=>log(68x+4)=log(3x+7)+log10 ('.' 1=log10;)

=>log(68x+4)=log((3x+7)*10)

=>log(68x+4)=log(30x+70)

=> 68x+4=30x+70

=>38x=66

=> x=66/38
=> x=33/19;

=>

Like? Yes (3) | No   

vinay chouhan   7 years AGO

log4+log(17x+1)-log(3x+7)=log{(68x+4)/(3x+7)}
0^0!=0 and 0^0=1
9 mod 11=9 and 243 mod 13=9
.'.log{(68x+4)/(3x+7)}=-0+1+9-9
log{(68x+4)/(3x+7)}=1
68x+4=10*(3x+7)
38x=66
x=66/38
x=33/19

Like? Yes | No   

Abhijit   7 years AGO

0^(0!)=0
0^(0)=1
9mod11=9
243mod13=9

Like? Yes | No