IIT JEE , AIEEE Forum - Mathematics , Algebra

Orgist

Hey guys, solve this please.

prove that,

log 1/2 to the base log ₃2 > 0

nadeemoidu

If a<1 then log_ab >0 when b<1           (1)
&
If a>1 then log_ab >0 when b>1

here a= log ₃2 and b= 1/2

2<3    =>    log₃2 <1

so from (1) , the result can be proven

spideyunlimited

PROOF :

log ₃2 = log 2 / log 3 = 0.301 / 0.477 = 0.631

Now, let a number x be defined such that :

0.631 ^x= 1/2

(6.31 / 10) ^x = 5 / 10

We know that as the power of a fraction increases, its value decreases.

Hence x will be some positive number.

0.631 ^x= 1/2

x = log _0.6311/2

x = log _{log 2 (base 3)} 1/2

So,

log _{log 2 (base 3)} 1 / 2 > 0

Orgist

Hey please tell me some easy understandable process, I am only in class XI.

spideyunlimited

arre see na
1/2 > (1/2)^2 > (1/2)^3
because
1/2 > 1/4 > 1/8
as the power of the fraction increases, the value decreases.
so as log 2 to the base 3 is a fraction greater than 1/2, it must be raised to a positive power so that it becomes 1/2.
like 1/3 ^2 = 1/9 where 1/3 is greater than 1/9.

after that i simply converted this statement into log equation.

pratikanand

easy way u have solved it, gaurav then y can't this fellow understand. for the one who asked: buddy don't get scared by this . u r gonna face a lot more...

Orgist

Oh all right

Now I have my process

log 1/2 to the base log ₃2

= log 1/2 / log(log ₃2)

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

= 0-log 2 / loglog 2 - loglog 3

= log 2 / loglog 3 - loglog 2

hence 3>2

loglog 3 > loglog2

therefore log 2 / loglog 3 - loglog 2 > 0

karthik_18049200

oh.very easy.
base is less than 1.
there fore,log _ax is a decreasing graph having (1,0)on x-axis.
un can observe that if x-co ordinate goes on increasing,the graph comes on negative y-axis.Means the value(required)becomes negative
so,if x is less than 1,the required value becomes +ve.
so,in the problem,1/2=x which is less than 1.
hence the solution.
cheers

karthik_18049200

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