MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
saumya shivhare Grade: 11
        

1. LOG(x-1)(base 4) = LOG(x-3)(base 2)  solution for 'x'


2. if logx(base 8) + log(x-4)(base 8) +log(x-6)(base 8) =2then x= ?


saumya shivhare

6 years ago

Answers : (2)

Aniket Patra
48 Points
										

Ans 1 is most probably 2 and 5.

6 years ago
palash ahuja
19 Points
										

ist question solution is here 


log(x-1)base4 = 1/2 log(x-1)(base2)


therefore 


1/2log(x-1)base2 = log(x-3)(base2)


therefore 


log(x-1)base2 = 2log(x-3)base2 = log(x-3)^2(base2)


since both sides have the same base therefore we can remove the log 


we get


(x-1)=(x-3)^2 = x^2 - 6x +9


on solving further we get


x^2-7x+10=0


x^2-2x-5x+10=0


x(x-2) -5(x-2)=0


(x-5)(x-2)=0


or x=5 or x=2 


 


iind solution is here


logx(base8)+log(x-4)base8+log(x-6)base8=2


then 


log x(x-4)(x-6)(base8)=2


or 


x(x-4)(x-6)=64


on solving we get a cubic equation


x^3 -10x^2 +24x - 64=0


let the roots of the given eq. be a,b and c


then a + b+c= 10


& abc = 64


now by a.m.-g.m. inequality we get


(a+b+c)> Or = 3(abc)^1/3 = 3*4 = 12


which is less than a+b+c 


hence no solutions are possible 


i hope you have understood the solutions 


PALASH AHUJA 


 

6 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: Rs. 15,900
  • View Details
Get extra Rs. 6,837 off
USE CODE: JYOTI
Get extra Rs. 1,003 off
USE CODE: JYOTI

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details