The Q is (1/5)^m (1/4)^18= 1/2(10)^35

=> 5 ^ -m * (2 ^ -2)^18 = (2 ^ -1 )* (10 ^ -35)

=> (5^-m) * (2^-36) = (2 ^ -1 )* (2 ^ -35) * (5 ^ -35 )

=> 5^-m = [ (2 ^ -36) * (5 ^ -35 )] / (2^-36) = (5 ^ -35 )

=> 5^-m = (5 ^ -35 )

=> m = 35

taneja.niks wrote:If (1/5)m(1/4)18 = 1/2(10)35, m?

on simplifying i get

(5^-35)(2^-35) = (2^35)(5^35)

how to proceed further???

Plz explain
ans m=35

You lost the m somewhere. Doesn't matter.

Shovan gave a possible explanation - I'd just like to point out the overall motive of solving exponential equations.
Whenever you have variables in the exponents, the solution is to rewrite the powers so that the bases are the same on both sides of the equation.
In the example above, you want to get everything on both sides to (5^something) * (2 ^something)

Once you have that, you can ignore the bases and equate the exponents. If 5^m*2^x equals 5*35*2^70, then
m=35
and
x=70.

Note that in the question above, you don't even need to mess around with the 2s and 4s, as only the 5s have variables in the exponents. Thus, the problem can be solved by breaking down 10 into 2*5, isolating (1/5)35 to have the same base as (1/5)^m on the left side, then, since the bases are the same, ignore the bases of 1/5 and equate the exponents: m=35.

shovan85 wrote:The Q is (1/5)^m (1/4)^18= 1/2(10)^35

=> 5 ^ -m * (2 ^ -2)^18 = (2 ^ -1 )* (10 ^ -35)

=> (5^-m) * (2^-36) = (2 ^ -1 )* (2 ^ -35) * (5 ^ -35 )

=> 5^-m = [ (2 ^ -36) * (5 ^ -35 )] / (2^-36) = (5 ^ -35 )

=> 5^-m = (5 ^ -35 )

=> m = 35

simply great yar ....thank u very much...