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

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If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink]

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16 Apr 2010, 17:41
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TOPIC LOCKED. FOR OPEN DISCUSSION OF THIS QUESTION, REFER TO if-1-5-m-1-4-18-frac-127321.html?fl=similar

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

A. 17
B. 18
C. 34
D. 35
E. 36
[Reveal] Spoiler: OA

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Re: If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink]

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17 Apr 2010, 07:17
shahany wrote:
I have a feeling I am missing something obvious, but I cannot figure out how to quickly calculate the answer to this:

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

a) 17
b) 18
c) 34
d) 35
3) 36

Thank you!

Long way:
$$(\frac{1}{5})^m*(\frac{1}{4})^{18}= \frac{1}{2*10^{35}}$$ --> $$\frac{1}{5^m}*\frac{1}{2^{36}}= \frac{1}{2*2^{35}*5^{35}}$$ --> $$\frac{1}{5^m}*\frac{1}{2^{36}}= \frac{1}{2^{36}*5^{35}}$$ --> $$\frac{1}{5^m}= \frac{1}{5^{35}}$$ --> $$m=35$$.

Short way:
$$(\frac{1}{5})^m * (\frac{1}{4})^{18} = \frac{1}{2*10^{35}}$$ --> $$\frac{1}{5^m}* (\frac{1}{4})^{18} = \frac{1}{2*2^{35}*5^{35}}$$ --> as there are only integers in the answer choices then we can concentrate only on the power of 5: they should be equal on both sides --> $$m=35$$.

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Re: If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink]

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17 Apr 2010, 18:50
Very helpful - thank you! I need to brush up on my properties...

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Re: If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink]

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14 May 2010, 06:49
vannbj wrote:
If $$(\frac{1}{5})^m (\frac{1}{4})^1^8 = \frac{1}{(2(10)^3^5}$$ then m =

A) 17
B) 18
C) 34
D) 35
E) 36

IMHO D

$$(1/5^m)$$ $$(1/2^36)$$ = 1/$$(2 * 2^35 * 2^35)$$
$$(1/5^m)$$ $$(1/2^36)$$ = 1/$$(2^36 * 2^35)$$
m= 35

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Re: If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink]

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14 May 2010, 10:38
indeed, m=35 , option D is the answer.

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Re: If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink]

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24 Jul 2010, 02:52
Ah, I got it.

rewrite for simplicity:
$$5^{-m} * 4^{-18} = 2^{-1} * 10^{-35}$$

Divide both sides by 2 (ie. multiply by 2^(-1)) and split up the term on the RHS:
$$2 * 5^{-m} * 2^{-18} * 2^{-18} = 2^{-35} * 5^{-35}$$

Group the term on the LHS:
$$2 * 5^{-m} * 2^{-36} = 2^{-35} * 5^{-35}$$

And divide both sides by 2^(-36):
$$2 * 5^{-m} = 2 * 5^{-35}$$

Divide both sides by 2, and you're left with:
$$5^{-m} = 5^{-35}$$

and so m = 35

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Re: If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink]

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09 Sep 2010, 15:31
Nice work, everyone - I posted this on another thread today, too, and thought it was fitting that the next question I clicked on would use the same ideology so here it is, also:

For exponents, your core competencies are:

Multiplication (so you should look to factor addition/subtraction to set up an opportunity to multiply)
Finding common bases for multiplication and equations (which is what you need to use here to get the 10 in terms of 2 and 5)
Looking for patterns and number properties (units digit, positive/negative, etc.)

When you see exponent-based problems, the above skills are those that you should be looking for, and keeping those at the forefront of your mind will help you to make quick, efficient decisions on how to proceed.
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Re: If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink]

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03 Oct 2010, 18:36
TehJay wrote:
This was the first problem I got on my practice exam today and it totally threw me off. I couldn't for the life of me figure it out. I'm sure I'm just forgetting basic exponent rules, but how do you do it?

$$(1/5^m)(1/{4^{18}}) = (1/(2*{10^{35}}))$$

$$(1/5^m)(1/{4^{18}}) = (1/(2^{1}*{2^{35} * 5^{35}}))$$

$$(1/5^m)(1/4^{18}) = (1/({2^{36}}*{5^{35}}))$$

Equating the powers of 5=> m=35 (D).
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Re: If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink]

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03 Oct 2010, 18:54
TehJay wrote:
This was the first problem I got on my practice exam today and it totally threw me off. I couldn't for the life of me figure it out. I'm sure I'm just forgetting basic exponent rules, but how do you do it?

So you are given that $$\frac{1}{5^m} * \frac{1}{4^1^8} = \frac{1}{2*10^3^5}$$

Breaking down the prime factors to the bare levels we get:

LHS = $$\frac{1}{5^m} * \frac{1}{4^1^8} = \frac{1}{5^m} * \frac{1}{2^3^6} = \frac{1}{5^m * 2^3^6}$$

RHS = $$\frac{1}{2*10^3^5} = \frac{1}{2* (5*2)^3^5} = \frac{1}{2^3^6 * 5^3^5}$$

Thus m = 35.

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Re: If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink]

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24 Oct 2010, 12:35
(1/5)^m (1/4)^18 = 1/2(10)^35

or, (1/5)^m (1/2)^36 =(1/2) * 1/(10)^35

or, (1/5)^m= 2^35 * {1/(10)^35}
or, (1/5)^m = (2/10)^35
or, (1/5)^m = (1/5)^35
Hence m =35.

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Re: If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink]

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24 Oct 2010, 12:41
Thank you for responding, so when you say "or" at the beginning of each line are you illustrating successive steps in solving the question or showing alternative ways to find the solution?
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Re: If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink]

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24 Oct 2010, 12:45
Given:

$$(\frac{1}{5})^m$$ * $$(\frac{1}{2})^(36)$$ = $$(\frac{1}{2})$$ * $$(\frac{1}{10})^(35))$$

Since, 10 = 2 * 5 and $$a^n$$ * $$b^n$$ = $$(ab)^n$$

==> $$(\frac{1}{5})^m$$ * $$(\frac{1}{2})^(36)$$ = $$(\frac{1}{2})$$ * $$(\frac{1}{2})^(35)$$ *$$(\frac{1}{5})^(35)$$

==> Cancelling $$(\frac{1}{2})^(36)$$ on both sides gives m = 35
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Re: If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink]

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24 Oct 2010, 12:55
Thank you! How do you cancel (1/2)^36 on both sides?
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Re: If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink]

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14 Nov 2010, 14:47
Pepe wrote:
Can you help me with following question?

If $$(\frac{1}{5})^m * (\frac{1}{4})^{18} = \frac{1}{2(10)^{35}$$, what value has m?

A 17
B 18
C 34
D 35
E 36

(1/4)^18 = 1/4^18 = 1/2^36

1/2*10^35 = 1/2*2^35*10^35 = 1/2^36*5^35

(1/5)^m = 1/5^m

Thus:

1/5^m*2^36 = 1/2^36*5^m

The correct answer is D m=35

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Re: If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink]

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Re: If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ?   [#permalink] 18 Jul 2014, 23:35
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