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Find Values of x...

Date: 9/4/96 at 22:33:30
From: michael atienza
Subject: Find Values of x...

Determine all real values of x for which
       2         x - 2x - 48
     (x - 5x + 5)             = 1


Date: 9/5/96 at 9:35:8
From: Doctor Jerry
Subject: Re: Find Values of x...

Dear Margarette,

You might consider for which real numbers a and b is a^b = 1 
(the carat (^) is often used to indicate powers).  Provided that 
a is not 0, a^0=1.  So, you could seek values of x for which 

If you know about logarithms, you can get another approach to the 
problem. In the equation a^b = 1, take logs of both sides to get
b*ln(a) = ln(1) = 0.  So, either b or ln(a) must be zero.

Hopefully, you can now finish the problem.  

-Doctor Jerry,  The Math Forum
 Check out our web site!   

Date: 9/5/96 at 18:39:15
From: m.atienza
Subject: Re: Find Values of x...

Unfortunately I am still stuck on this problem.

I factored the exponent (x - 2x - 48) and came up with
  x=8 and x=-6                                              2
The thing I don't understand is what do I do with the base x  - 5x + 5 
with regards to the exponent that I factored? Will there be more than 
one value for x?

Thanks for your help!


Date: 9/5/96 at 20:49:22
From: Doctor Robert
Subject: Re: Find Values of x...

x^2 - 5x + 5 is the base, x^2-2x-48 is the exponent, and 1 is the 
result, if I read your problem correctly.  Now, the only way that a 
number raised to a power can be 1 is if the exponent is zero, that is, 
z^0 = 1.  In your problem this means that x^2-2x-48 = 0.  You solve 
this by factoring

(x-8)(x+6) = 0 

which leads to the conclusion that x=8 or x=-6.  

The only thing you need to check is that these values of x do not make 
the base zero, for zero raised to the zero power is not defined.  

-Doctor Robert,  The Math Forum
 Check out our web site!   

Date: 09/09/2000 at 02:15:07
From: Tim Greene
Subject: Re: Find Values of x...

In the solution above, the writer only finds the solutions

     x = -6   and   x = 8 
by finding solutions of the form a^0 = 1 (a not = 0). There are three 
other solutions to the problem.

Finding solutions of the form 1^a = 1 (no restrictions on a), we have

     (x^2 - 5x + 5) = 1 

which gives the two solutions:

     x = 1   and   x = 4.  
Finding solutions of the form (-1)^a = 1 (where a is an even integer), 
we have:
     (x^2 - 5x + 5) = -1

this gives the two potential solutions:

     x = 2   and   x = 3

however, for x = 2 the exponent (x^2 - 2x - 48) is an even integer but 
for x = 3 that exponent is odd.

Thus, there are five solutions in all: -6, 1, 2, 4, and 8.

Dr. Greenie
Associated Topics:
High School Basic Algebra
High School Polynomials

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