FACTORING AND EXPONENTS!!!

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