EricImasogie wrote:Hey guys, I need your help in understanding this one.

for the right side of the equation I read online that essentially you can pull out (2^35)*(5^35) which equates to 10^35 . I am having trouble conceptualizing this because I always thought that (2)(5^35) was equal to 10^35.

(ab)^35 = a^35 * b^35. Think about what you said Eric. How could 2^1 * 5^35 possibly equal 10^35? That's not logical. You can't just double 5^35 and get 10^35.

Check this out. 10^5 = (5 * 2)^5. So we need five 5's and five 2's to get 10^5.

10^5 = 10 * 10 * 10 * 10 * 10 = (5 * 2) * (5 * 2) * (5 * 2) * (5 * 2) * (5 * 2) = 5 * 5 * 5 * 5 * 5 * 2 * 2 * 2 * 2 * 2.

In the problem, the denominator in the right side of the equation has 35 5's and 35 + 1 = 36 2's.

In the denominator on the left side, 4^18 = (2 * 2)^18 = 2^36. So we are all set with enough 2's. Now to get enough 5's we need m to equal 35.

Choose D.

If (1/5)^m * (1/4)^18 = 1/2(10)^35 then m = ?

A. 17

B. 18

C. 34

D. 35

E. 36

No need to simplify the whole equation.

Every prime factor on the right side must also appear on the left side.
On the right side we have 1/10³� = 1/2³� * 1/5³�.
Thus, the left side must also include 1/5³�, implying that m=35.

The correct answer is D.

(1/5)^m * (1/4)^18 = 1/(2(10)^35)
Find the reciprocal:
5^m * 4^18 =2(10)^35
5^m * 4^18 =2 x 2^35 x 5^35)
5^m * 4^18 =2 x 2^17 x 2^18 x 5^35
5^m * 4^18 =2^36 x 5^35
5^m * 4^18 =4^18 x 5^35
m = 35

Mathsbuddy wrote:(1/5)^m * (1/4)^18 = 1/(2(10)^35)
Find the reciprocal:
5^m * 4^18 =2(10)^35
5^m * 4^18 =2 x 2^35 x 5^35)
5^m * 4^18 =2 x 2^17 x 2^18 x 5^35
5^m * 4^18 =2^36 x 5^35
5^m * 4^18 =4^18 x 5^35
m = 35

Or put more simply, the only exponent of 5 is 35.

Let's cross multiply to get rid of the denominators.

This gives us 2*10³� = 5� * 4¹�, or

2*2³�*5³� = 5� * 2³�

Hence m = 35, and we're done!