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				   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

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### Answers : (2)

										Ans 1 is most probably 2 and 5.

6 years ago
										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

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