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Thanks for answering my past answer people.

show work for solving this one please:

5^x^2-4x+3= 1

I'm getting just x=3?

2006-10-26 08:06:28 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

4 answers

Write 1 as 5^0.

Then, since the bases are equal, the exponents must also be equal, so x^2 - 4x + 3 = 0.

Factor: (x-1)(x-3) = 0

so x=1 or x=3.

2006-10-26 08:09:03 · answer #1 · answered by James L 5 · 4 1

the expression is:
5^(x^2 -4x + 3) = 1
RHS can be written as 5^0. This is done to have LHS and RHS to have the same base
5^(x^2 - 4x + 3) = 5^0
since the bases are same, The powers can be compared
hence (x^2 -4x + 3) = 0
Factorizing
(x - 3)(x - 1) = 0
Equating both factors to zero successively, we obtain
x = 3 and x = 1

2006-10-26 16:08:02 · answer #2 · answered by quark_sa 2 · 0 0

5^(x^2-4x+3)=5^0
since the base is the samethe powers can be compared
=>x^2-4x+3=0
=>(x-3)(x-1)=0
x=1 or 3

2006-10-26 15:17:19 · answer #3 · answered by raj 7 · 0 2

ok
if you want to solve:
5^(x^2-4x+3) = 1
then you need to find:
(x^2-4x+3) = 0
and this can be factorized as:
(x-3)(x-1)= 0
so there are 2 solutions:
x=3 and x=1

2006-10-26 15:19:56 · answer #4 · answered by locuaz 7 · 0 1

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