Can we write?

Question

If
1^(2n+3) = 1^(3n+2)
can we write
2n+3 = 3n+2?

Answer 1

Hello Rashej

Yes we can write that we can prove it by contradiction
Suppose that our statement is not correct
thus although 1^(2n+3) = 1^(3n+2)
2n+3 - 3n+2 is not 0, for some n
thus for some n , -n +5 , is not 0.

For n = 1 it is not 0 but clearly 1^(5) = 1^(5)

This contradicts our assumtion and thus the negation of our assumtion is valid.

Answer 2

No! Here's why.
1 creates problems when indices and exponents are used. Here's an example to show you why it isn't proper to write 2n+3 = 3n+2:
1^0 = 1
1^125 = 1
So, 1^0 = 1^125
Does that mean 0 =125?
Except when n = 1, you can't write 2n+3 = 3n+2

Answer 3

No because 1 at any power equals 1. If instead of 1, there was something else, you could have written 2n+3=3n+2, that means n=1.

Answer 4

If you read the answer on your previous answer you should know the answer ont his one.

IF 1^(2n+3) = 1^(3m+2) THEN the answer is NO

IF 1^(2n+3) = 1^(3n+2) THEN the answer is YES

Answer 5

it is true for all bases except 0.1 and -1 considering real base.

because 0^n = 0; 1^n= 1 and (-1) ^ n = 1 or -1

considering n being integer,

Things become complex is n is not integer as 1^(1/3) = 1, w, w^2 where w is complex cube root of one. In this case it does not hold

Answer 6

Not at all.

This is because 1^n = 1 for -∞ < n < ∞. This means you cannot say, because 1^6 = 1^255, then 6=255.

Such expressions fail at 1, 0, and -1.

Answer 7

yes it can be written in that way as
a^n =a^m then
n=m
In this case value of n = 1

Answer 8

naaaaaa, coz anything raised to the power 1 is 1 only

Answer 9

no bcoz if n is -ve then LHS is not equal to RHS and this equation is valid only when n=1 for greater values of n..again LHS is not equal to RHS

Answer 10

I dont see what the problem is with that equation. I think it is possible in that way.