I'm studying for a test and ran into a tough math question, (1/5)^m * (1/4)^18 = 1/(2(10)^35)?

Question

I'm supossed to solve for M, I know the answer is 35, but I'm not sure how to get there. Can somone explain the steps for solving this problem. Thanks

Answer 1

Let's write the equation as follows:

[1 / (5)^m] * [1 / (4)^18] = 1 / [2* (10^35)]

We can reduce it to:
1 / [(5)^m * (4)^18] = 1 / [2* (10^35)]

Let's get rid of the ugly factions by cross-multiplying:
2*(10^35) = 5^m * 4^18

Now we have to notice that we can split 4^18 into (2*2)^18, and then reduce that to 2^36.

In the same way, we notice that 10^35 can be split as (2*5)^35, or 2^35 * 5^35.

Let's rewrite as follows:
2 * 2^35 * 5^35 = 5^m * 2^36

Let's combine and we see:
2^36 * 5^35 = 5^m * 2^36. The 2^36 cancels out of both sides and we get

5^35 = 5^m, and therefore m=35 (just by visual inspection).

Well technically you have to reduce a little further, but I'm not sure if you need to introduce any algebra functions in your solution, but the more technical solution would proceed as follows:
5^35 = 5^m. [take the log of both sides - using whichever base you want, log-10, ln, lg-2, or whatever you like - I'll write log to be generic].
log (5^35) = log (5^m) [the exponent comes down in front when you take the log of a power]
35 log(5) = m log(5) [log(5) cancels out]
35=m [and that's your solution].

Answer 2

(1/5)^m * (1/4)^18 = 1 / [2(10^35)]

1/(5^m) * 1/(4^18) = 1/[2(10^35)]

Cross-multiply:

(5^m) (4^18) = 2(10^35)

(5^m) (2^36) = 2(10^35)

Take log of both sides:

m log 5 + 36 log 2 = log 2 + 35 log 10

m log 5 = 35 log 10 - 35 log 2 = 35(log 10 - log 2) = 35 log 5

m = 35

Answer 3

YOu need a very good calc like ti-83
first divide both sides by (1/4)^18
now you have (1/5)^m = 3.43...e-25
now convert it to a log problem
log base(.2) 3.43xe-25 = m
you have to use the2nd ans key to get the exact values out of calculator
now since your calc does not do base .2 youhave to change it to base 10
log(2nd ans)/log(.2)
and out pops the number 35

Answer 4

bear in concepts that the version of squares continually components right into a manufactured from sum and variations: x^2 - y^2 = (x+y)*(x-y). truthfully, i think of all of those are meant to be the version of squares: are you advantageous the 2d isn't a^2 - 4b^2? then the climate would be (a+2b)*(a-2b). The third is (7y + 13z)*(7y - 13z). study to renowned appropriate squares in this variety of subject (distinction of appropriate squares): 4b^2 = (2b)^2 49y^2 = (7y)^2 169z^2 = (13z)^2