Answer:
x > 1/2
Step-by-step explanation:
The lower limint of the domain of the log function, regardless of the base, is x > 0. But that's not all. If the right side is 2, then there's an upper limit to x.
If the solution is x = 2, then we have:
log 4.5x = 2
2(2)-1
or...
log 9 = 2. This is a true statement, since 9 = 3^2.
3
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The question here is whether or not there's an upper limit on x.
You claim that the solution is x=2. Is that the upper limit of the domain? Let's find out. Let's try x = 3:
log 13.5
5 = 2 iS THIS TRUE OR FALSE?
Use the change of base formula:
log 13.5 = log 13.5 / log 5 = 1.617. This is actually less than 2
5.
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Let me play the devil's advocate and hypothesize for a moment that the solution is not x = 2. If not 2, then what is x?
(2x - 1) ^ log (4.5x) = (2x -1)^2, or 4x^2 - 4x + 1
2x -1
Then 4.5x = 4x^2 - 4x + 1, or
4x^2 - 8.5x + 1 = 0
This is a quadratic function, with a = 4, b = - 8.5 and c = 1. Then the discriminant is b^2 - 4(a)(c), which here is (-8.5)^2 - 4(4)(1) ), or 56.25.
This would lead to roots of
- (-8.5) plus or minus √56.25
x = ---------------------------------------------------
2(4)
x = 2 or x = 1/8 (You claim that the solution is x = 2).
Let's try once more to identify the domain:
(1) log (4.5x) = 2 is valid only when the base, 2x-1, is greater than
2x-1 zero. Setting 2x - 1 > 0, we get 2x ≥ 1. or x > 1/2. For values of x less than 1/2, the base (2x - 1) would be 0 or smaller, which is not permissible when working with logs.
This domain, x > 1/2, applies only to the function log y
2x-1
You claim that x = 2. Does that satisfy x > 1/2, the domain of x? Yes. So that's promising.
Based upon these explorations, I'd claim that the domain of the log part of this equation is x > 1/2.
