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If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?

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If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?  [#permalink]

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New post 15 Feb 2020, 17:58
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A
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C
D
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Re: If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?  [#permalink]

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New post Updated on: 15 Feb 2020, 18:05
If \(a^m*a^n = a^{mn}\), what is the value of \(m(n-2)+n(m-2)\)?
A. -1
B. -1/2
C. 0
D. ½
E. 2

\(a^m*a^n = a^{mn}\) => \(a^{m+n} = a^{mn}\)

=> m+n = mn

\(m(n-2)+n(m-2) = mn-2m+nm-2n = 2(mn-m-n) = 2(mn-(m+n)) = 0\) - Answer C

Originally posted by shameekv1989 on 15 Feb 2020, 18:04.
Last edited by shameekv1989 on 15 Feb 2020, 18:05, edited 1 time in total.
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Re: If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?  [#permalink]

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New post 15 Feb 2020, 18:04
a^m*a^n = a^mn
a^m*a^n = a^m+n
Hence m+n=mn
m(n-2)+n(m-2)
=2mn - 2(m+n)
=2mn- 2mn
=0

C is the answer

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Re: If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?  [#permalink]

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New post Updated on: 15 Feb 2020, 20:31
From the first formula we can find m+n=mn by simplifying the equation as did below.
a^m*a^n=a^(m+n)
a^(m+n)=a^mn => m+n=mm

Then we try to simplify the problem as did below
m(n-2)+n(m-2)=mn-2m+mn-2m=2mn-2m-2n=2(mn-m-n)=2(mn-(m+n))
As you can see the simplified outcome is 2(mn-(m+n))

Since m+n=mm we can substitute m+n for mn as did below
2(mn-(m+n)=2(mn-mn)=O
So C is right

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Originally posted by Enkhhulan on 15 Feb 2020, 18:14.
Last edited by Enkhhulan on 15 Feb 2020, 20:31, edited 1 time in total.
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Re: If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?  [#permalink]

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New post 15 Feb 2020, 18:25
a^m.a^n=a^mn
a^(m+n)=a^mn
m+n=mn
m(n-2)+n(m-2)
=mn-2m+mn-2n
=2mn-2(m+n)
=2mn-2mn
=0
Hence, Ans. is C.
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Re: If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?  [#permalink]

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New post 15 Feb 2020, 18:41
The answer would be 0. This is a pretty easy one.
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Re: If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?  [#permalink]

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New post 15 Feb 2020, 20:05
given
mn = m+n

equation
m(n-2)+n(m-2)
mn-2m +mn-2n
2mn-2(m+n)
from given stmt
2mn -2mn =0
thus C
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Re: If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?  [#permalink]

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New post 15 Feb 2020, 21:35
Solution:

Question 1: If \(a^m*a^n = a^{mn}\), what is the value of \(m(n-2)+n(m-2)\)?


LHS = \(a^m*a^n = a^{m + n}\)

RHS = \(a^{mn} \)

By equating both the LHS and RHS, we get: \(a^{m + n} = a^{mn}\) [Equation 1]

Now, in order to find out the value of expression, \(m(n - 2) + n(m - 2)\), we would first expand the expression by removing the parenthesis.

Therefore, \(m(n - 2) + n(m - 2) = mn - 2m + mn - 2n = 2mn - 2(m + n) = 2(mn - m - n)\) [Equation 2]

From Equation 1 we get, \(m + n = mn\) since, base is ''a'' on LHS and RHS.

Thus, \(mn - m - n = 0 \) and by substituting this value in Equation 2 we get: \(2(mn - m - n) = 2*0 = 0\), which is the answer.

Therefore, Option C i.e., 0 is the correct answer of question 1.
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Re: If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?  [#permalink]

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New post 15 Feb 2020, 23:07
If a^m*a^n = a^mn, what is the value of m(n-2)+n(m-2)?

a^m*a^n = a^mn

a^(m+n) = a^mn ---(laws of exponent)

from here we get -- (m+n) = mn
now,
m(n-2)+n(m-2) = mn-2m+mn-2n = 2mn-2(m+n) = 2{mn-(m+n)} = 0 (since m+n = mn)

C is the correct answer
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Re: If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?  [#permalink]

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New post 15 Feb 2020, 23:17
a^m*a^n = a^mn
a^{m+n} = a^mn
m+n = mn

let m=2, n=2
Substituting values of m and n in the expression m(n-2)+n(m-2) = 0

IMO C
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Re: If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?  [#permalink]

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New post Updated on: 16 Feb 2020, 06:21
GIVEN: \( a^{m} * a^{n} = a^{mn}\)
TO FIND: m(n-2)+n(m-2)

given: \( a^{m} * a^{n} = a^{mn}\)
=> \( a^{m+n} = a^{mn}\)
=> \(m+n = mn \)
=> \(m + n - mn = 0\) --- Eqn. (1)

We have to find the value of:
= \( m(n-2)+n(m-2) \)
= \( mn - 2m + mn - 2n \)
= \( 2mn - 2m - 2n \)
= \(2(mn - m - n)\) --- Eqn. (2)
= \( -2 (m + n - mn) \) {taking -1 as common from the bracket}
= \( -2 ( 0 ) = 0 \) {as we know from Eqn. (1) that: m + n - mn = 0}

Hence, our answer is 0.

Originally posted by asterias on 15 Feb 2020, 23:50.
Last edited by asterias on 16 Feb 2020, 06:21, edited 1 time in total.
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Re: If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?  [#permalink]

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New post 16 Feb 2020, 00:10
a^m.a^n=a^mn

a^(m+n)=a^mn

m+n=mn

Let it be equation 1

Now

m(n-2)+n(m-2)

=mn-2m+mn-2n

=2mn-2(m+n)

And from equation 1 (m+n)=mn

=2mn-2mn

=0
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Re: If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?  [#permalink]

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New post 16 Feb 2020, 02:29
upon solving m(n-2)+n(m-2) we get 2(mn-m-n) =>2(mn - (m+n)) => 2(mn-mn) => ans zero (C)
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Re: If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?  [#permalink]

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Re: If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?  [#permalink]

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New post 16 Feb 2020, 05:20
Solution:

a^m*a^n = a^mn => m+n=mn => mn-m-n = 0

for the question asked:

value of mn-2m+mn-2n => 2(mn-m-n) => 0 since mn-m-n = 0;


Answer : 0
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Re: If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?  [#permalink]

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New post 16 Feb 2020, 07:26
OK lets do it in within 1 minute.

From the given equation :mn=m+n

-->2(mn-(m+n))=0
So
Ans: (C) 0
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Re: If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?   [#permalink] 16 Feb 2020, 07:26

If a^m*a^n = a^(mn), what is the value of m(n-2)+n(m-2)?

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