Hi Chris

The way to do these kind of questions is to first figure out how the base numbers in both sides of the equations are related. In this case, 2 and 5 are a factor of 10. i.e 10 can be written as 2 x 5

Now the power to which 10 is raised is the same as the powers to which 2 and 5 are raised.

so if (1/10)^35 = (1/ (2 x 5))^35

Now, according to the question given,

(1/5)^m (1/4)^18 = 1/2(10)^35
= (1/5)^m (1/2)^36 = 1/2(10)^35
Now, equating the powers of the bases
= (1/5)^m (1/2)^36 = 1/2(2 x 5)^35
= (1/5)^m (1/2)^36 = 1/2 x (1/10)^35
= (1/5)^m (1/2)^36 = 1/2 x (1/2 x 5)^35
= (1/5)^m (1/2)^36 = 1/2 x (1/2)^35 x (1/5)^35
= 1/2 x (1/5)^m x (1/2)35 = 1/2 x (1/2)^35 x (1/5)^35
Which shows that m = 35

Now for teh second one
If 2^x - 2^(x-2) = 3(2^13) - (1)

let 2^x be y, then substituting in equation (1), we get

Also, as a^-2 = (1/a)^2, 2^(x-2) = 2^x X 2^-2 => y X (1/2)^2 => y/4

y - y/4 = 3(2^13)
3y/4 = 3(2^13)
y/4 = 2^13
y = 2^13 X 2^2
y = 2^15

y = 2^15 = 2^x
Thus x =15

Hope this helps
GyanOne

How did you get from 1/2(2 x 5)^35 to 1/2 x (1/10)^35

hi chris...
for the second question i can suggest u much simpler way..
2^x-2^(x-2)=3(2^13)
we take teh L.H.S
2^x-(2^x*2^-2)
taking common 2^x
2^x[(1-2^-2)]
2^x[(1-1/4)]
2^x[(3/4)]
2^x[3/(2^2)]
3[2^(x-2)]

comparing this with the R.H.S
we get,
x-2=13
that implies
x=15