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 2 x - 2x - 48 (x - 5x + 5) = 1 Thanks, Margarette 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 x^2-2x-48=0. 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! http://mathforum.org/dr.math/ 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. 2 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! Margarette 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! http://mathforum.org/dr.math/ 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 |
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